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Question Number 221661    Answers: 1   Comments: 1

Question Number 221647    Answers: 0   Comments: 1

solve for x. x^1 + x^2 + x^3 = 4096

$${solve}\:{for}\:{x}. \\ $$$${x}^{\mathrm{1}} \:+\:{x}^{\mathrm{2}} \:+\:{x}^{\mathrm{3}} \:\:=\:\:\mathrm{4096} \\ $$

Question Number 221638    Answers: 1   Comments: 0

Question Number 221637    Answers: 1   Comments: 0

Question Number 221626    Answers: 3   Comments: 0

Question Number 221620    Answers: 1   Comments: 0

Solve for x ((7x))^(1/3) =(√x)[x≠0]

$${Solve}\:{for}\:{x} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{7}{x}}=\sqrt{{x}}\left[{x}\neq\mathrm{0}\right] \\ $$

Question Number 221618    Answers: 3   Comments: 0

solve for x 2^x +4^x =8^x

$${solve}\:{for}\:{x} \\ $$$$\mathrm{2}^{{x}} +\mathrm{4}^{{x}} =\mathrm{8}^{{x}} \\ $$

Question Number 221669    Answers: 2   Comments: 3

Question Number 221601    Answers: 2   Comments: 0

∫ ((sin 2x)/(1 + sin 3x)) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\frac{{sin}\:\mathrm{2}{x}}{\mathrm{1}\:+\:{sin}\:\mathrm{3}{x}}\:{dx} \\ $$$$ \\ $$

Question Number 221592    Answers: 2   Comments: 0

Question Number 221588    Answers: 1   Comments: 0

∫ ((8t − 8t^( 3) )/(t^( 6) + 6t^5 + 3t^( 4) − 20t^3 + 3t^2 + 6t + 1)) dt

$$ \\ $$$$\:\:\:\:\int\:\frac{\mathrm{8}{t}\:−\:\mathrm{8}{t}^{\:\mathrm{3}} }{{t}^{\:\mathrm{6}} \:+\:\mathrm{6}{t}^{\mathrm{5}} \:+\:\mathrm{3}{t}^{\:\mathrm{4}} \:−\:\mathrm{20}{t}^{\mathrm{3}} \:+\:\mathrm{3}{t}^{\mathrm{2}} \:+\:\mathrm{6}{t}\:+\:\mathrm{1}}\:{dt}\:\:\:\: \\ $$$$ \\ $$

Question Number 221587    Answers: 1   Comments: 0

∫_( 2) ^( 3) ((tan^(− 1) (x))/(1 − x^2 )) dx

$$\int_{\:\mathrm{2}} ^{\:\mathrm{3}} \:\frac{\mathrm{tan}^{−\:\mathrm{1}} \left(\mathrm{x}\right)}{\mathrm{1}\:\:−\:\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$

Question Number 221586    Answers: 1   Comments: 0

∫ ((sin 2x)/(1 + 3x )) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{1}\:+\:\mathrm{3}{x}\:}\:{dx} \\ $$$$ \\ $$

Question Number 221585    Answers: 6   Comments: 1

solve for x ∈R (x^3 −6)^3 =x+6

$${solve}\:{for}\:{x}\:\in{R} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{6}\right)^{\mathrm{3}} ={x}+\mathrm{6} \\ $$

Question Number 221583    Answers: 1   Comments: 0

Question Number 221578    Answers: 1   Comments: 0

$$\:\:\: \\ $$

Question Number 221576    Answers: 0   Comments: 0

Question Number 221577    Answers: 1   Comments: 0

Prove; ∫_0 ^( 1) ((xdx)/((x^2 + 1)(e^(2πx) − 1))) = (γ/2) − (1/4) where; γ is a Euler′s Mascheroni constant

$$ \\ $$$$\:\:\:\:\:\:\mathrm{Prove};\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}\mathrm{d}{x}}{\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({e}^{\mathrm{2}\pi{x}} \:−\:\mathrm{1}\right)}\:=\:\frac{\gamma}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:\: \\ $$$$\:\:\:\:\mathrm{where};\:\gamma\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{Mascheroni}\:\mathrm{constant}\:\:\:\: \\ $$$$ \\ $$

Question Number 221582    Answers: 0   Comments: 0

(1):∫_0 ^u x^(ν−1) (u^2 −x^2 )^(ϱ−1) e^(μx) dx=Σ_(n=0) ^∞ (μ^n /(n!))∫_0 ^u x^(ν+n−1) (u^2 −x^2 )^(ϱ−1) dx =Σ_(n=0) ^∞ (μ^n /(n!))∙(u^(2ϱ+ν+n−2) /2)B(((ν+n)/2),ϱ) =(u^(2ϱ+ν+2) /2)Σ_(n=0) ^∞ (((μν)^n )/(n!)) ((Γ(((ν+n)/2))Γ(ϱ))/(Γ(((ν+n)/2)+ϱ))) =((u^(2ρ+ν−2) Γ(ϱ))/2)[Σ_(k=0) ^∞ (((μν)^(2k) )/((2k)!)) ((Γ(ν/2+k))/(Γ(ν/2+ϱ+k)))+Σ_(k=0) ^∞ (((μν)^(2k+1) )/((2k+1))) ((Γ(((ν+1)/2)+k))/(Γ(((ν−1)/2)+ϱ+k)))] =((u^(2ϱ+ν−2) Γ(ϱ))/2)[((Γ(ν/2))/(Γ(ν/2+ϱ))) _1 F_2 ((ν/2);(1/2),(ν/2)+ϱ;((μ^2 u^2 )/4))+((μνΓ(((ν+1)/2)))/(Γ(((ν+1)/2)+ϱ))) _1 F_2 (((ν+1)/2);(3/2),((ν+1)/2)+ϱ;((μ^2 u^2 )/4))] =(1/( 2))B(ν,ϱ)u^(2ν+2ϱ+2) _1 F_2 (ν;(1/2)+ϱ,((μ^2 u^2 )/4))+(μ/2)B(ν+(1/2),ϱ)u^(2ν+2ϱ−1) _1 F_2 (ν+(1/2);(3/2),ν+ϱ+(1/2);((μ^2 u^2 )/4)) (2):=Σ_(k=0) ^∞ (μ^k /(k!))∫_0 ^1 x^(ν+k−1) (u^2 −x^2 )^(ϱ−1) dx =^(x=ut) u^(ν+2ϱ−1) Σ_(k=0) ^∞ (((μu)^k )/(k!))∫_0 ^1 t^(ν+k−1) (1−t^2 )^(ϱ−1) dt =(u^(ν+2ϱ−1) /2)Σ_(k=0) ^∞ (((μν)^k )/(k!))B(((ν+k)/2),ϱ) =(u^(ν+2ϱ−1) /2)[Σ_(m=0) ^∞ (((μν)^(2m) )/((2m)!))B(ν+m,ϱ)+Σ_(m=0) ^∞ (((μu)^(2m+1) )/((2m+1)))B(ν+m+(1/2),ϱ)] =((B(ν,ϱ))/2)u^(ν+2ϱ−1) Σ_(m=0) ^∞ (((ν)_m )/((ν+ϱ)_m )) (((μν/2)^(2m) )/(m!)) +((μB(ν+(1/2),ϱ))/2)u^(ν+2ϱ) Σ_(m=0) ^∞ (((ν+(1/2))_m )/((ν+ϱ+(1/2))_m )) (((μν/2)^(2m) )/((2m+1)!!)) =(1/( 2))B(ν,ϱ)u^(2ν+2ϱ+2) _1 F_2 (ν;(1/2)+ϱ,((μ^2 u^2 )/4))+(μ/2)B(ν+(1/2),ϱ)u^(2ν+2ϱ−1) _1 F_2 (ν+(1/2);(3/2),ν+ϱ+(1/2);((μ^2 u^2 )/4)) (3):∫_0 ^u x^(ν−1) (u^2 −x^2 )^(ϱ−1) e^(μx) dx=(u^(2ϱ+ν−2) /2)Γ(ϱ)Γ((ν/2))Σ_(k=0) ^∞ (((((μ^2 u^2 )/4))^k )/(k!Γ((ν/2)+k+ϱ)Γ(k+(1/2)))) Γ(k+(1/2))=(((2k−1)!!)/2^k )(√π),Γ((ν/2)+k+ϱ)=Γ((ν/2)+ϱ)((ν/2)+ϱ)_k Σ_(k=0) ^∞ (z^k /(k!((ν/2)+ϱ)_k Γ(k+(1/2))))=(1/( Γ(1/2)))Σ_(k=0) ^∞ (((1)_k z^k )/(((ν/2)+ϱ)_k k!((1/2))_k ))=(1/( (√π))) _1 F_2 (1;(1/2),(ν/2)+ϱ;z) _1 F_2 (1;(1/2),(ν/2)+ϱ;z)=Σ_(k=0) ^∞ (((1)_k z^k )/(((1/2))_k ((ν/2)+ϱ)_k k!)) z=((μ^2 u^2 )/4),(1)_k =k! _1 F_2 (1;(1/2),(ν/2)+ϱ;((μ^2 +ν^2 )/4))=Σ_(k=0) ^∞ (((((μ^2 −u^2 )/4))^k )/(((1/2))_k ((ν/2)+ϱ)_k )) =(1/( 2))B(ν,ϱ)u^(2ν+2ϱ+2) _1 F_2 (ν;(1/2)+ϱ,((μ^2 u^2 )/4))+(μ/2)B(ν+(1/2),ϱ)u^(2ν+2ϱ−1) _1 F_2 (ν+(1/2);(3/2),ν+ϱ+(1/2);((μ^2 u^2 )/4)) (4):∫_0 ^u x^(ν−1) (u^2 −x^2 )^(ϱ−1) e^(μx) dx=∫_0 ^u x^(ν−1) (u^2 −x^2 )^(ϱ−1) e^(μx) dx x=ut⇒dx=udt,t∈(0,1) =∫_0 ^1 (ut)^(ν−1) (u^2 (1−t^2 ))^(ϱ−1) e^(μut) udt=u^(ν−1) u^(2ϱ−2) u∫_0 ^1 t^(ν−1) (1−t^2 )^(ϱ−1) e^(μut) dt =u^(v+2ρ−2) ∫_0 ^1 t^(ν−1) (1−t^2 )^(ϱ−1) e^(μut) dt e^(μνt) =Σ_(k=0) ^∞ (((μut)^k )/(k!)) =u^(ν+2ϱ−2) Σ_(k=0) ^∞ (((μu)^k )/(k!))∫_0 ^1 t^(ν+k−1) (1−t^2 )^(ϱ−1) dt t^2 =s⇒t=s^(1/2) ,dt=(1/2)s^(−1/2) ds ∫_0 ^1 t^(ν+k−1) (1−t^2 )^(ϱ−1) dt=∫_0 ^1 s^(((ν+k)/2)−(1/2)) (1−s)^(ϱ−1) ds=(1/2)B(((ν+k)/2),ϱ) =(1/2)u^(ν+2ϱ−2) Σ_(k=0) ^∞ (((μu)^k )/(k!))B(((ν+k)/2),ϱ) B(a,b)=((Γ(a)Γ(b))/(Γ(a+b))) =(1/2)u^(ν+2ϱ−2) Γ(ϱ)Σ_(k=0) ^∞ (((μu)^k )/(k!)) ((Γ(((ν+k)/2)))/(Γ(((ν+k)/2)+ϱ))) k=2m ∧ k=2m+1 =(1/2)u^(ν+2ϱ−2) Γ(a)[Σ_(m=0) ^∞ (((μu)^(2m) )/((2m)!)) ((Γ((ν/2)+m))/(Γ((ν/2)+m+ϱ)))+Σ_(m=0) ^∞ (((μu)^(2m+1) )/((2m+1)!)) ((Γ(((ν+1)/2)+m))/(Γ(((ν−1)/2)+m+ϱ)))] Γ(a+m)=Γ(a)(a)_m =(1/2)u^(ν+2ϱ−2) Γ(a)[((Γ((ν/2)))/(Γ((ν/2)+ϱ)))Σ_(m=0) ^∞ (((μu)^(2m) )/((2m)!))((ν/2))_m /((ν/2)+ϱ)_m +((Γ(((ν+1)/2)))/(Γ(((ν+1)/2)+ϱ)))Σ_(m=0) ^∞ ((((μu)^(2m+1) )/((2m+1)!)) ((Γ(((ν+1)/2)+m))/(Γ(((ν+1)/2)+m+ϱ)))] (1/((2m)!))=(1/(4^m ((1/2))_m m!)),(1/((2m+1)!))=(1/(2^(2m) ((3/2))_m m!))∙(1/(2m+1))∙(1/(Γ((3/2))))Γ((3/2)) (2m)!=4^m ((1/2))_m m!,(2m+1)(2m)!=(2m+1)^(4m) ((1/2))_m m!=2^(2m+1) ((m!)/((−))) ((3/2))=((Γ(m+(3/2)))/(Γ((3/2))))=(((2m+1)!!)/2^m ) (2m+1)!=(2m+1)!!∙2^m m!=(((2m+1)!)/(2^m m!))∙2^m m!=(2m+1)! Σ_(m=0) ^∞ (((μu)^(2m) )/((2m)!)) ((((ν/n))_m )/(((ν/2)+ϱ)_m ))=Σ_(m=0) ^∞ (((((μ^2 u^2 )/4))^m )/(((1/2))_m m!)) ((((ν/2))_m )/(((1/2))_m )) _1 F_2 ((ν/2);(1/2),(ν/2)+ϱ;((μ^2 u^2 )/4)) Σ_(m=0) ^∞ (((μn)^(2m+1) )/((2m)!)) ((((ν/2))_m )/(((ν/2)+ϱ)))= _1 F_2 ((ν/2);(1/2),(ν/2)+ϱ;((μ^2 u^2 )/4)) Σ_(m=0) ^∞ (((μu)^(2m+1) )/((2m+1)!)) (((((ν+1)/2))_m )/(Γ(((ν+1)/2)+ϱ)_m ))=(μu)Σ_(m=0) ^∞ (((μ^2 u^2 )^m )/((2m+1)!)) (((((ν+1)/2))_m )/((((ν+1)/2)+ϱ)))=μu _1 F_2 (((ν+1)/2);(3/2),((ν+1)/2)+ϱ;((μ^2 u^2 )/4)) B((ν/2),ϱ)=((Γ((ν/2))Γ(ϱ))/(Γ((ν/2)+ϱ))),B(((ν+1)/2),ϱ)=((Γ(((ν+1)/2))Γ(ϱ))/(Γ(((ν+1)/2)+ϱ))) Γ(ϱ)=((Γ((ν/2)))/(Γ((ν/2)+ϱ)))=((Γ((ν/2))Γ(ϱ))/(Γ((ν/2)+ϱ)))=B((ν/2),ϱ)=B((ν/2),ϱ) Γ(ϱ)((Γ((ν/2)))/(Γ((ν/2)+ϱ)))=B((ν/2),ϱ) Γ(ϱ)((Γ(((ν+1)/2)))/(Γ(((ν+1)/2)+ϱ)))=B(((ν+1)/2),ϱ) =(1/2)u^(ν+2ϱ−2) [B((ν/2),ϱ)Σ_(m=0) ^∞ (((μu)^(2m) )/((2m)!)) ((((ν/2))_m )/(((ν/2)+ϱ)_m ))+B(((ν+1)/2),ϱ)Σ_(m=0) ^∞ (((μu)^(2m+1) )/((2m+1)!)) (((((ν+1)/2))_m )/((((ν+1)/2)+ϱ)_m ))] Σ_(m=0) ^∞ (((μu)^(2m) )/((2m)!)) ((((v/2))_m )/(((ν/2)+ϱ)))= _1 F_2 ((ν/2);(1/2),(ν/2)+ϱ;((μ^2 u^2 )/4)) Σ_(m=0) ^∞ (((μu)^(2m+1) )/((2m+1)!)) (((((ν+1)/2))_m )/((((ν+1)/2)+ϱ)_m ))=μu _1 F_2 (((ν+1)/2);(1/2);(ν/2)+ϱ;((μ^2 u^2 )/4)) B((ν/2),ϱ)=((Γ((ν/2))Γ(ϱ))/(Γ((ν/2)+ϱ))),B(((ν+1)/2),ϱ)=((Γ(((ν+1)/2))Γ(ϱ))/(Γ(((ν+1)/2)+ϱ))) (1/2)u^(ν+2ϱ−2) B((ν/2),ϱ) _1 F_2 ((ν/2);(1/2),(ν/2)+ϱ;((μ^2 u^2 )/4))+(1/2)u^(ν+2ϱ−2) ∙μuB(((ν+1)/2),ϱ) _1 F_(2 ) (((ν+1)/2);(3/2),((ν+1)/2)+ϱ;((μ^2 u^2 )/4)) =(1/2)B((ν/2),ϱ)u^(ν+2ϱ−2) _1 F_2 ((ν/2);(1/2),(ν/2)+ϱ;((μ^2 u^2 )/4))+(μ/2)B(((ν+1)/2),ϱ)u^(ν+2ϱ−1) _1 F_2 (((ν+1)/2);(3/2),((ν+1)/2)+ϱ;((μ^2 u^2 )/4)) =(1/( 2))B(ν,ϱ)u^(2ν+2ϱ+2) _1 F_2 (ν;(1/2)+ϱ,((μ^2 u^2 )/4))+(μ/2)B(ν+(1/2),ϱ)u^(2ν+2ϱ−1) _1 F_2 (ν+(1/2);(3/2),ν+ϱ+(1/2);((μ^2 u^2 )/4))

$$\left(\mathrm{1}\right):\int_{\mathrm{0}} ^{{u}} {x}^{\nu−\mathrm{1}} \left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {e}^{\mu{x}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mu^{{n}} }{{n}!}\int_{\mathrm{0}} ^{{u}} {x}^{\nu+{n}−\mathrm{1}} \left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {dx} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mu^{{n}} }{{n}!}\centerdot\frac{{u}^{\mathrm{2}\varrho+\nu+{n}−\mathrm{2}} }{\mathrm{2}}{B}\left(\frac{\nu+{n}}{\mathrm{2}},\varrho\right) \\ $$$$=\frac{{u}^{\mathrm{2}\varrho+\nu+\mathrm{2}} }{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu\nu\right)^{{n}} }{{n}!}\:\frac{\Gamma\left(\frac{\nu+{n}}{\mathrm{2}}\right)\Gamma\left(\varrho\right)}{\Gamma\left(\frac{\nu+{n}}{\mathrm{2}}+\varrho\right)} \\ $$$$=\frac{{u}^{\mathrm{2}\rho+\nu−\mathrm{2}} \Gamma\left(\varrho\right)}{\mathrm{2}}\left[\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu\nu\right)^{\mathrm{2}{k}} }{\left(\mathrm{2}{k}\right)!}\:\frac{\Gamma\left(\nu/\mathrm{2}+{k}\right)}{\Gamma\left(\nu/\mathrm{2}+\varrho+{k}\right)}+\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu\nu\right)^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)}\:\frac{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+{k}\right)}{\Gamma\left(\frac{\nu−\mathrm{1}}{\mathrm{2}}+\varrho+{k}\right)}\right] \\ $$$$=\frac{{u}^{\mathrm{2}\varrho+\nu−\mathrm{2}} \Gamma\left(\varrho\right)}{\mathrm{2}}\left[\frac{\Gamma\left(\nu/\mathrm{2}\right)}{\Gamma\left(\nu/\mathrm{2}+\varrho\right)}\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}},\frac{\nu}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{\mu\nu\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho\right)}\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu+\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}},\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}}{B}\left(\nu,\varrho\right){u}^{\mathrm{2}\nu+\mathrm{2}\varrho+\mathrm{2}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\nu;\frac{\mathrm{1}}{\mathrm{2}}+\varrho,\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{\mu}{\mathrm{2}}{B}\left(\nu+\frac{\mathrm{1}}{\mathrm{2}},\varrho\right){u}^{\mathrm{2}\nu+\mathrm{2}\varrho−\mathrm{1}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\nu+\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}},\nu+\varrho+\frac{\mathrm{1}}{\mathrm{2}};\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\left(\mathrm{2}\right):=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mu^{{k}} }{{k}!}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\nu+{k}−\mathrm{1}} \left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {dx} \\ $$$$\overset{{x}={ut}} {=}{u}^{\nu+\mathrm{2}\varrho−\mathrm{1}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{{k}} }{{k}!}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\nu+{k}−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {dt} \\ $$$$=\frac{{u}^{\nu+\mathrm{2}\varrho−\mathrm{1}} }{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu\nu\right)^{{k}} }{{k}!}{B}\left(\frac{\nu+{k}}{\mathrm{2}},\varrho\right) \\ $$$$=\frac{{u}^{\nu+\mathrm{2}\varrho−\mathrm{1}} }{\mathrm{2}}\left[\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu\nu\right)^{\mathrm{2}{m}} }{\left(\mathrm{2}{m}\right)!}{B}\left(\nu+{m},\varrho\right)+\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}+\mathrm{1}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)}{B}\left(\nu+{m}+\frac{\mathrm{1}}{\mathrm{2}},\varrho\right)\right] \\ $$$$=\frac{{B}\left(\nu,\varrho\right)}{\mathrm{2}}{u}^{\nu+\mathrm{2}\varrho−\mathrm{1}} \underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\nu\right)_{{m}} }{\left(\nu+\varrho\right)_{{m}} }\:\frac{\left(\mu\nu/\mathrm{2}\right)^{\mathrm{2}{m}} }{{m}!} \\ $$$$+\frac{\mu{B}\left(\nu+\frac{\mathrm{1}}{\mathrm{2}},\varrho\right)}{\mathrm{2}}{u}^{\nu+\mathrm{2}\varrho} \underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\nu+\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} }{\left(\nu+\varrho+\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} }\:\frac{\left(\mu\nu/\mathrm{2}\right)^{\mathrm{2}{m}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)!!} \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}}{B}\left(\nu,\varrho\right){u}^{\mathrm{2}\nu+\mathrm{2}\varrho+\mathrm{2}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\nu;\frac{\mathrm{1}}{\mathrm{2}}+\varrho,\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{\mu}{\mathrm{2}}{B}\left(\nu+\frac{\mathrm{1}}{\mathrm{2}},\varrho\right){u}^{\mathrm{2}\nu+\mathrm{2}\varrho−\mathrm{1}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\nu+\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}},\nu+\varrho+\frac{\mathrm{1}}{\mathrm{2}};\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\left(\mathrm{3}\right):\int_{\mathrm{0}} ^{{u}} {x}^{\nu−\mathrm{1}} \left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {e}^{\mu{x}} {dx}=\frac{{u}^{\mathrm{2}\varrho+\nu−\mathrm{2}} }{\mathrm{2}}\Gamma\left(\varrho\right)\Gamma\left(\frac{\nu}{\mathrm{2}}\right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)^{{k}} }{{k}!\Gamma\left(\frac{\nu}{\mathrm{2}}+{k}+\varrho\right)\Gamma\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\Gamma\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)!!}{\mathrm{2}^{{k}} }\sqrt{\pi},\Gamma\left(\frac{\nu}{\mathrm{2}}+{k}+\varrho\right)=\Gamma\left(\frac{\nu}{\mathrm{2}}+\varrho\right)\left(\frac{\nu}{\mathrm{2}}+\varrho\right)_{{k}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{{k}} }{{k}!\left(\frac{\nu}{\mathrm{2}}+\varrho\right)_{{k}} \Gamma\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\:\Gamma\left(\mathrm{1}/\mathrm{2}\right)}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}\right)_{{k}} {z}^{{k}} }{\left(\frac{\nu}{\mathrm{2}}+\varrho\right)_{{k}} {k}!\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }=\frac{\mathrm{1}}{\:\sqrt{\pi}}\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\mathrm{1};\frac{\mathrm{1}}{\mathrm{2}},\frac{\nu}{\mathrm{2}}+\varrho;{z}\right) \\ $$$$\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\mathrm{1};\frac{\mathrm{1}}{\mathrm{2}},\frac{\nu}{\mathrm{2}}+\varrho;{z}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}\right)_{{k}} {z}^{{k}} }{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} \left(\frac{\nu}{\mathrm{2}}+\varrho\right)_{{k}} {k}!} \\ $$$${z}=\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}},\left(\mathrm{1}\right)_{{k}} ={k}! \\ $$$$\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\mathrm{1};\frac{\mathrm{1}}{\mathrm{2}},\frac{\nu}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} +\nu^{\mathrm{2}} }{\mathrm{4}}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mu^{\mathrm{2}} −{u}^{\mathrm{2}} }{\mathrm{4}}\right)^{{k}} }{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} \left(\frac{\nu}{\mathrm{2}}+\varrho\right)_{{k}} } \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}}{B}\left(\nu,\varrho\right){u}^{\mathrm{2}\nu+\mathrm{2}\varrho+\mathrm{2}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\nu;\frac{\mathrm{1}}{\mathrm{2}}+\varrho,\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{\mu}{\mathrm{2}}{B}\left(\nu+\frac{\mathrm{1}}{\mathrm{2}},\varrho\right){u}^{\mathrm{2}\nu+\mathrm{2}\varrho−\mathrm{1}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\nu+\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}},\nu+\varrho+\frac{\mathrm{1}}{\mathrm{2}};\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\left(\mathrm{4}\right):\int_{\mathrm{0}} ^{{u}} {x}^{\nu−\mathrm{1}} \left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {e}^{\mu{x}} {dx}=\int_{\mathrm{0}} ^{{u}} {x}^{\nu−\mathrm{1}} \left({u}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {e}^{\mu{x}} {dx} \\ $$$${x}={ut}\Rightarrow{dx}={udt},{t}\in\left(\mathrm{0},\mathrm{1}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left({ut}\right)^{\nu−\mathrm{1}} \left({u}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)\right)^{\varrho−\mathrm{1}} {e}^{\mu{ut}} {udt}={u}^{\nu−\mathrm{1}} {u}^{\mathrm{2}\varrho−\mathrm{2}} {u}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\nu−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {e}^{\mu{ut}} {dt} \\ $$$$={u}^{{v}+\mathrm{2}\rho−\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\nu−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {e}^{\mu{ut}} {dt} \\ $$$${e}^{\mu\nu{t}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{ut}\right)^{{k}} }{{k}!} \\ $$$$={u}^{\nu+\mathrm{2}\varrho−\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{{k}} }{{k}!}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\nu+{k}−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {dt} \\ $$$${t}^{\mathrm{2}} ={s}\Rightarrow{t}={s}^{\mathrm{1}/\mathrm{2}} ,{dt}=\frac{\mathrm{1}}{\mathrm{2}}{s}^{−\mathrm{1}/\mathrm{2}} {ds} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\nu+{k}−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\varrho−\mathrm{1}} {dt}=\int_{\mathrm{0}} ^{\mathrm{1}} {s}^{\frac{\nu+{k}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{s}\right)^{\varrho−\mathrm{1}} {ds}=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\nu+{k}}{\mathrm{2}},\varrho\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\nu+\mathrm{2}\varrho−\mathrm{2}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{{k}} }{{k}!}{B}\left(\frac{\nu+{k}}{\mathrm{2}},\varrho\right) \\ $$$${B}\left({a},{b}\right)=\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\nu+\mathrm{2}\varrho−\mathrm{2}} \Gamma\left(\varrho\right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{{k}} }{{k}!}\:\frac{\Gamma\left(\frac{\nu+{k}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\nu+{k}}{\mathrm{2}}+\varrho\right)} \\ $$$${k}=\mathrm{2}{m}\:\wedge\:{k}=\mathrm{2}{m}+\mathrm{1} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\nu+\mathrm{2}\varrho−\mathrm{2}} \Gamma\left({a}\right)\left[\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}} }{\left(\mathrm{2}{m}\right)!}\:\frac{\Gamma\left(\frac{\nu}{\mathrm{2}}+{m}\right)}{\Gamma\left(\frac{\nu}{\mathrm{2}}+{m}+\varrho\right)}+\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}+\mathrm{1}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)!}\:\frac{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+{m}\right)}{\Gamma\left(\frac{\nu−\mathrm{1}}{\mathrm{2}}+{m}+\varrho\right)}\right] \\ $$$$\Gamma\left({a}+{m}\right)=\Gamma\left({a}\right)\left({a}\right)_{{m}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\nu+\mathrm{2}\varrho−\mathrm{2}} \Gamma\left({a}\right)\left[\frac{\Gamma\left(\frac{\nu}{\mathrm{2}}\right)}{\Gamma\left(\frac{\nu}{\mathrm{2}}+\varrho\right)}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}} }{\left(\mathrm{2}{m}\right)!}\left(\frac{\nu}{\mathrm{2}}\right)_{{m}} /\left(\frac{\nu}{\mathrm{2}}+\varrho\right)_{{m}} +\frac{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho\right)}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(\mu{u}\right)^{\mathrm{2}{m}+\mathrm{1}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)!}\:\frac{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+{m}\right)}{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+{m}+\varrho\right)}\right]\right. \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{2}{m}\right)!}=\frac{\mathrm{1}}{\mathrm{4}^{{m}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} {m}!},\frac{\mathrm{1}}{\left(\mathrm{2}{m}+\mathrm{1}\right)!}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{m}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{m}} {m}!}\centerdot\frac{\mathrm{1}}{\mathrm{2}{m}+\mathrm{1}}\centerdot\frac{\mathrm{1}}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2}{m}\right)!=\mathrm{4}^{{m}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} {m}!,\left(\mathrm{2}{m}+\mathrm{1}\right)\left(\mathrm{2}{m}\right)!=\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{4}{m}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} {m}!=\mathrm{2}^{\mathrm{2}{m}+\mathrm{1}} \frac{{m}!}{\left(−\right)} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\Gamma\left({m}+\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}=\frac{\left(\mathrm{2}{m}+\mathrm{1}\right)!!}{\mathrm{2}^{{m}} } \\ $$$$\left(\mathrm{2}{m}+\mathrm{1}\right)!=\left(\mathrm{2}{m}+\mathrm{1}\right)!!\centerdot\mathrm{2}^{{m}} {m}!=\frac{\left(\mathrm{2}{m}+\mathrm{1}\right)!}{\mathrm{2}^{{m}} {m}!}\centerdot\mathrm{2}^{{m}} {m}!=\left(\mathrm{2}{m}+\mathrm{1}\right)! \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}} }{\left(\mathrm{2}{m}\right)!}\:\frac{\left(\frac{\nu}{{n}}\right)_{{m}} }{\left(\frac{\nu}{\mathrm{2}}+\varrho\right)_{{m}} }=\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)^{{m}} }{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} {m}!}\:\frac{\left(\frac{\nu}{\mathrm{2}}\right)_{{m}} }{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} }\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}},\frac{\nu}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{n}\right)^{\mathrm{2}{m}+\mathrm{1}} }{\left(\mathrm{2}{m}\right)!}\:\frac{\left(\frac{\nu}{\mathrm{2}}\right)_{{m}} }{\left(\frac{\nu}{\mathrm{2}}+\varrho\right)}=\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}},\frac{\nu}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}+\mathrm{1}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)!}\:\frac{\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}\right)_{{m}} }{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho\right)_{{m}} }=\left(\mu{u}\right)\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu^{\mathrm{2}} {u}^{\mathrm{2}} \right)^{{m}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)!}\:\frac{\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}\right)_{{m}} }{\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho\right)}=\mu{u}\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu+\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}},\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$${B}\left(\frac{\nu}{\mathrm{2}},\varrho\right)=\frac{\Gamma\left(\frac{\nu}{\mathrm{2}}\right)\Gamma\left(\varrho\right)}{\Gamma\left(\frac{\nu}{\mathrm{2}}+\varrho\right)},{B}\left(\frac{\nu+\mathrm{1}}{\mathrm{2}},\varrho\right)=\frac{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\varrho\right)}{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho\right)} \\ $$$$\Gamma\left(\varrho\right)=\frac{\Gamma\left(\frac{\nu}{\mathrm{2}}\right)}{\Gamma\left(\frac{\nu}{\mathrm{2}}+\varrho\right)}=\frac{\Gamma\left(\frac{\nu}{\mathrm{2}}\right)\Gamma\left(\varrho\right)}{\Gamma\left(\frac{\nu}{\mathrm{2}}+\varrho\right)}={B}\left(\frac{\nu}{\mathrm{2}},\varrho\right)={B}\left(\frac{\nu}{\mathrm{2}},\varrho\right) \\ $$$$\Gamma\left(\varrho\right)\frac{\Gamma\left(\frac{\nu}{\mathrm{2}}\right)}{\Gamma\left(\frac{\nu}{\mathrm{2}}+\varrho\right)}={B}\left(\frac{\nu}{\mathrm{2}},\varrho\right) \\ $$$$\Gamma\left(\varrho\right)\frac{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho\right)}={B}\left(\frac{\nu+\mathrm{1}}{\mathrm{2}},\varrho\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{u}^{\nu+\mathrm{2}\varrho−\mathrm{2}} \left[{B}\left(\frac{\nu}{\mathrm{2}},\varrho\right)\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}} }{\left(\mathrm{2}{m}\right)!}\:\frac{\left(\frac{\nu}{\mathrm{2}}\right)_{{m}} }{\left(\frac{\nu}{\mathrm{2}}+\varrho\right)_{{m}} }+{B}\left(\frac{\nu+\mathrm{1}}{\mathrm{2}},\varrho\right)\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}+\mathrm{1}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)!}\:\frac{\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}\right)_{{m}} }{\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho\right)_{{m}} }\right] \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}} }{\left(\mathrm{2}{m}\right)!}\:\frac{\left(\frac{{v}}{\mathrm{2}}\right)_{{m}} }{\left(\frac{\nu}{\mathrm{2}}+\varrho\right)}=\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}},\frac{\nu}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mu{u}\right)^{\mathrm{2}{m}+\mathrm{1}} }{\left(\mathrm{2}{m}+\mathrm{1}\right)!}\:\frac{\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}\right)_{{m}} }{\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho\right)_{{m}} }=\mu{u}\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu+\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};\frac{\nu}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$${B}\left(\frac{\nu}{\mathrm{2}},\varrho\right)=\frac{\Gamma\left(\frac{\nu}{\mathrm{2}}\right)\Gamma\left(\varrho\right)}{\Gamma\left(\frac{\nu}{\mathrm{2}}+\varrho\right)},{B}\left(\frac{\nu+\mathrm{1}}{\mathrm{2}},\varrho\right)=\frac{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\varrho\right)}{\Gamma\left(\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{u}^{\nu+\mathrm{2}\varrho−\mathrm{2}} {B}\left(\frac{\nu}{\mathrm{2}},\varrho\right)\:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}},\frac{\nu}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{u}^{\nu+\mathrm{2}\varrho−\mathrm{2}} \centerdot\mu{uB}\left(\frac{\nu+\mathrm{1}}{\mathrm{2}},\varrho\right)\:_{\mathrm{1}} {F}_{\mathrm{2}\:} \left(\frac{\nu+\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}},\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\nu}{\mathrm{2}},\varrho\right){u}^{\nu+\mathrm{2}\varrho−\mathrm{2}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}},\frac{\nu}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{\mu}{\mathrm{2}}{B}\left(\frac{\nu+\mathrm{1}}{\mathrm{2}},\varrho\right){u}^{\nu+\mathrm{2}\varrho−\mathrm{1}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\frac{\nu+\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}},\frac{\nu+\mathrm{1}}{\mathrm{2}}+\varrho;\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$=\frac{\mathrm{1}}{\:\mathrm{2}}{B}\left(\nu,\varrho\right){u}^{\mathrm{2}\nu+\mathrm{2}\varrho+\mathrm{2}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\nu;\frac{\mathrm{1}}{\mathrm{2}}+\varrho,\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right)+\frac{\mu}{\mathrm{2}}{B}\left(\nu+\frac{\mathrm{1}}{\mathrm{2}},\varrho\right){u}^{\mathrm{2}\nu+\mathrm{2}\varrho−\mathrm{1}} \:_{\mathrm{1}} {F}_{\mathrm{2}} \left(\nu+\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}},\nu+\varrho+\frac{\mathrm{1}}{\mathrm{2}};\frac{\mu^{\mathrm{2}} {u}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$

Question Number 221554    Answers: 0   Comments: 0

Question Number 221550    Answers: 0   Comments: 0

∫_0 ^(π/2) ((x((tan x))^(1/4) )/(sin x))dx=(((3−2(√2)))/(24))Γ((1/8))Γ((3/8)) (1):I=∫_0 ^(π/2) ((x tan^(1/4) x)/(sin x))dx=∫_0 ^(π/2) x tan^(−3/4) x sec x dx tan x=t⇒dx=(dt/(1+t^2 )),sec x=(√(1+t^2 )) I=∫_0 ^∞ ((arctant∙t^(−3/4) )/( (√(1+t^2 ))))dt arctan t=∫_0 ^1 (t/(1+y^2 t^2 ))dy I=∫_0 ^1 ∫_0 ^∞ (t^(1/4) /((1+y^2 t^2 )(√(1+t^2 ))))dt dy t=(u/y)⇒dt=(du/y),t^2 =(u^2 /y^2 ) I=∫_0 ^1 y^(−5/4) ∫_0 ^∞ (u^(1/4) /((1+u^2 )(√(1+(u^2 /y^2 )))))du dy (√(1+(u^2 /y^2 )))=(√((y^2 +u^2 )/y)) I=∫_0 ^1 y^(−1/4) ∫_0 ^∞ (u^(1/4) /((1+u^2 )(√(y^2 +u^2 ))))du dy ∫_0 ^∞ (u^(1/4) /((1+u^2 )(√(y^2 +u^2 ))))dx=(π/(2(√(y^2 −1))))(y^(−1/4) −y^(5/4) ) I=(π/2)∫_0 ^1 ((y^(−1/4) −y^(−5/4) )/( (√(y^2 −1))))dy y=sinθ⇒dy=cos θ dθ I=(π/2)∫_0 ^(π/2) ((sin^(−1/4) θ−sin^(5/4) θ)/( (√(sin θ−1))))cos θ dθ (√(sin^2 θ−1))=i cos θ I=(π/(2i))∫_0 ^(π/2) (sin^(−1/4) θ−sin^(−5/4) θ)dθ I=(π/(2i))∫_0 ^(π/2) (sin^(−1/4) −sin^(−5/4) θ)dθ ∫_0 ^(π/2) sin^a θ dθ=(((√π)Γ(((a+1)/2)))/(2Γ((a/2)+1))) I=(π/(2i))[(((√π)Γ((3/8)))/(2Γ((7/8))))−(((√π)Γ(−(1/8)))/(2Γ((3/8))))] Γ((7/8))Γ((1/8))=(π/(sin(π/8))),Γ(−(1/8))=−(8/7)Γ((7/8)) I=(π^(3/2) /(4i))[((Γ((3/8))Γ((1/8)))/π)sin(π/8)+((8Γ((7/8))Γ((3/8)))/(7Γ((3/8))))] sin(π/8)=((√(2−(√2)))/2) I=(π^(3/2) /(4i))[((Γ((3/8))Γ((1/8)))/π)∙((√(2−(√2)))/2)+((8Γ((7/8)))/7)] Γ((7/8))=(π/(Γ((1/8))sin(π/8))) I=(π^(3/2) /(4i))[((Γ((3/8))Γ((1/8)))/π)∙((√(2−(√2)))/2)+((8π)/(7Γ((1/8))sin(π/8)))] Γ((3/8))Γ((5/4))=(π/(sin((3π)/8))),sin((3π)/8)=((√(2+(√2)))/2) =(((3−2(√2)))/(24))Γ((1/8))Γ((3/8)) (2):u=tan x,du=(1+u^2 )dx,x=arctan u sin x=(u/( (√(1+u^2 )))),(((tan x))^(1/4) /(sin x))=(u^(1/4) /(u/( (√(1+u^2 )))))=u^(−(3/4)) (1+u^2 )^(1/2) dx=(du/(1+u^2 )),x∈[0,(π/2)]⇒x∈[0,∞] ∫_0 ^(π/2) ((x((tan x))^(1/4) )/(sin x))dx=∫_0 ^∞ arctan u∙u^(−(3/4)) (1+u^2 )^(−(1/2)) (du/(1+u^2 )) (1+u^2 )^(−(1/2)) (1+u^2 )^(−1) =(1+u^2 )^(−(3/2)) arctan u=∫_0 ^1 (u/(1+v^2 u^2 ))du ∫_0 ^∞ ∫_0 ^1 ((u∙u^(−(3/4)) (1+u^2 )^(−(1/2)) )/(1+v^2 u^2 ))dvdu=∫_0 ^1 ∫_0 ^∞ ((u^(1/4) (1+u^2 )^(−(1/2)) )/(1+v^2 u^2 ))dudv w=u^2 ,du=(1/2)w^(−(1/2)) dw ∫_0 ^∞ (((w^(1/2) )^(1/4) (1+w)^(−(1/2)) )/(1+v^2 w))∙(1/2)w^(−(1/2)) dw=(1/2)∫_0 ^∞ w^((1/5)−(1/2)−(1/2)) (1+w)^(−(1/2)) (1+v^2 w)^(−1) dw=(1/2)∫_0 ^∞ w^(−(7/8)) (1+w)^(−(1/2)) (1+v^2 w)^(−1) dw t=(w/(1+w)),w=(t/(1−t)),dw=(dt/((1−t)^2 )),w=0⇒t=0,w→∞⇒t−1^− 1+w=(1/(1−t)),1+v^2 w=1+v^2 (t/(1−t))=((1−t+v^2 t)/(1−t))=((1+(v^2 −1)t)/(1−t)) w^(−(7/8)) =((t/(1−t)))^(−(7/8)) =t^(−(7/8)) (1−t)^(7/8) (1+w)^(−(1/2)) =((1/(1−t)))^(−(1/2)) =(1−t)^(1/2) (1+v^2 w)^(−1) =(((1+(v^2 +1)t)/(1−t)))^(−1) =(1−t)(1+(v^2 −1)t)^(−1) dw=(dt/((1−t)^2 )) w^(−(7/8)) (1+w)^(−(1/2)) (1+v^2 w)^(−1) dw=t^(−(7/8)) (1−t)^(7/8) (1−t)^(1/2) (1+(v^2 −1)t)^(−1) (1−t)(dt/((1−t)^2 ))=t^(−(7/8)) (1−t)^((7/8)+(1/2)+1−2) (1+(v^2 −1)t)^(−1) dt (7/8)+(1/2)+1−2=(7/8)+(4/8)+(8/8)−((16)/8)=(3/8) t^(−(7/8)) =t^(−(1−(1/8))) =t^(−((8/8)−(1/8))) =t^(−(8/8)+1) =t^(−(((b−1)/b))) b−1=(7/8),b=1+(7/8)=((15)/8) b=(5/8),c−b=(7/8),c=b+(7/8)=(5/8)+(7/8)=((12)/8)=(3/2),a=1,z=−(v^2 −1)=1−v^2 ∫_0 ^1 t^(b−1) (1−t)^((c−b)−1) (1−zt)^(−a) dt=B(b,c−b) _2 F_1 (a,b;c;z) ∫_0 ^1 t^((5/8)−1) (1−t)^((7/8)−1) (1+(v^2 +1)t)^(−1) dt=B((5/8),(7/8))F_1 (1,(5/8);(3/2);1−v^2 ) B((5/8),(7/8))=((Γ((5/8))Γ((7/8)))/(Γ((3/2)))),Γ((3/2))=(1/2)Γ((1/2))=((√π)/2) (1/2)∫_0 ^∞ w^(−(7/8)) (1+w)^(−(1/2)) (1+v^2 w)^(−1) dw=(1/2)∙((Γ((5/8))Γ((7/8)))/((√π)/2)) _2 F_1 (1,(5/8);(3/2);1−v^2 )=((Γ((5/8))Γ((7/8)))/( (√π))) _2 F_1 (1,(5/8);(3/2);1−v^2 ) I=∫_0 ^1 ((Γ((5/8))Γ((7/8)))/( (√π))) _2 F_1 (1,(5/8);(3/2);1−v^2 )dv _2 F_1 (1,(5/8);(3/2);1−v^2 )dv=∫_0 ^1 Σ_(n=0) ^∞ (((1)_n ((5/8))_n )/(((3/2))_n n!))(1−v^2 ) ∫_0 ^1 _2 F_1 (1,(5/8);(3/2);1−v^2 )=Σ_(n=0) ^∞ (((1)_n ((5/8))_n )/(((3/2))_n n!))(1−v^2 )^n ∫_0 ^1 _2 F_1 (1,(5/8);(3/2);1−v^2 )=∫_(0 ) ^1 Σ_(n=0) ^∞ ((((5/8))_n )/(((3/2))_n n!))(1−v^2 )^n dv s=v^2 ,dv=(ds/(2s^(1/2) )),v:0→1⇒s:0→1 ∫_0 ^1 (1−s)^n (ds/(2s^(1/2) ))=(1/2)∫_0 ^1 s^(−(1/2)) (1−s)^n ds=(1/2)B((1/2),n+1)=(1/2) ((Γ((1/2))Γ(n+1))/(Γ(n+(3/2)))) Γ((1/2))=(√π),Γ(n+1)=n! Γ(n+(3/2))=Γ((n+1)+(1/2))=(((2n−1)!!)/2^(n+1) )(√π)=(((2n+1)!)/(2^(n+1) n!))(√π) B((1/2),n+1)=∫_0 ^1 s^((1/2)−1(1−s)^((n−1)−1) ds=((Γ((1/2))n+1)/(Γ(n+(3/2))))=(((√π)n!)/((((2n+1)!)/(2^(2n+1) n!))(√π)))=(((n!)^2 4^n Γ((1/2)))/(Γ(n+(1/2)+1)))) Γ(n+(3/2))=(((2n−1))/2^(n+1) )=(((2n−1)!!)/2^(n+1) )=(√π)=(((2n+1)!)/(2^(n+1) n!))(√π) B((1/2),n+1)=((Γ((1/2))Γ(n+1))/(Γ(n+(3/2))))=(((√π)n)/((((2n+1))/(2^(2n+1) n!))(√π))) (1/2)B((1/2),n+1)=(1/2) (((n!)^2 2^(2n+1) )/((2n+1)!))=(((n!)^2 2^(2n) )/((2n+1)!)) (2n+1)!=(2n+1)(2n)! (((n!)^2 2^(2n) )/((2n+1)!))=(((n!)^2 4^n )/((2n+1)(2n)!)) (1/2)B((1/2),n+1)=((Γ((1/2))Γ(n+1))/(2Γ(n+(3/2))))=(((√π)n!)/(2∙(((2n+1)!!)/2^(n+1) )(√π)))=(((n!)^2 4^n )/( (((2n)),(n) )(2n+1)n!)) ∫_0 ^1 (1−v^2 )^n dv=(1/2)B((1/2),n+1)=(1/2) ((Γ((1/2))Γ(n+1))/(Γ(n+(3/2))))=(1/2) (((√π)n!)/(Γ(n+(3/2)))) Γ(n+(3/2))=(((2n+1)!!)/2^(n+1) )(√π)=(((2n−1)!)/(4^n (n!)^2 2^(n+1) ))Γ((1/2))(√π) (1/2) (((√π)n!)/((((2n+1)!!)/2^(n+1) )(√π)))=(1/2) ((n!2^(n+1) )/((2n+1)!!))=((2^n n!)/((2n+1)!!)) (2n+1)!!=(((2n+1)!)/(2^n n!)) ((2^n n!)/((2n+1)!!))=((2^n n!)/(((2n+1)!)/(2^n n!)))=((4^n (n!)^2 )/((2n+1)!)) ∫_0 ^1 _2 F_1 (1,(5/8);(3/2);1−v^2 )dv=Σ_(n=0) ^∞ ((((5/8))_n )/(((3/2))_n n!))∫_0 ^1 (1−v^2 )^n dv=Σ_(n=0) ^∞ ((((5/8))_n )/(((3/2))_n n!))∙((4^n (n!)^2 )/((2n+1)!)) =Σ_(n=0) ^∞ ((((5/8))_n )/(((3/2))_n n!))∙((4^n n!)/((2n+1)!))∙(1/(n!)) (2n+1)!=(2n+1)(2n!), (((2n)),(n) )=(((2n)!)/((n!)^2 )),((4^n n!)/((2n)!))=(1/ (((2n)),(n) )) ((4^n n!)/((2n+1)!))=((4^n n!)/((2n+1)(2n)!))=(1/((2n+1) (((2n)),(n) ))) Σ_(n=0) ^∞ ((((5/8))_n )/(((3/2))_n n!))∙((4^n (n!)^2 )/((2n+1)!))=Σ_(n=0) ^∞ ((((5/8))_n (1)_n ((1/2))_n )/(((3/2))_n (1)_n n!))∙(1/((2n+1) (((2n)),(n) ))) I=((Γ((5/8))Γ((7/8)))/( (√π)))Σ_(n=0) ^∞ ((((5/8))_n (1)_n ((1/2))_n )/(((3/2))_n (1)_n n!))∙(1/((2n+1) (((2n)),(n) ))) Γ((5/8))Γ((7/8))=Γ(1−(3/8))Γ(1−(1/8))=(π/(sin(((3π)/8))))∙(π/(sin((π/8)))) sin(((3π)/8))=sin(π−((5π)/8))=sin(((3π)/8)),sin((π/8))=sin((π/8)) sin(((3π)/8))sin((π/8))=(1/2)(cos((π/4))−cos((π/2)))=(1/2)cos((π/4))=(1/2)∙((√2)/2) Γ((5/8))Γ((7/8))=(π^2 /(sin(((3π)/8))sin((π/8))))=(π^2 /((√2)/4))=((4π^2 )/( (√2)))=2(√2)π^2 ∫_0 ^(π/2) ((x((tan x))^(1/4) )/(sin x))dx=(((3−(√2))(√π))/(24))Γ((1/8))Γ((3/8))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}}}{\mathrm{sin}\:{x}}{dx}=\frac{\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{24}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right) \\ $$$$\left(\mathrm{1}\right):{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\:\mathrm{tan}^{\mathrm{1}/\mathrm{4}} {x}}{\mathrm{sin}\:{x}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\:\mathrm{tan}^{−\mathrm{3}/\mathrm{4}} {x}\:\mathrm{sec}\:{x}\:{dx} \\ $$$$\mathrm{tan}\:{x}={t}\Rightarrow{dx}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} },\mathrm{sec}\:{x}=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{arctan}{t}\centerdot{t}^{−\mathrm{3}/\mathrm{4}} }{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt} \\ $$$$\mathrm{arctan}\:{t}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}}{\mathrm{1}+{y}^{\mathrm{2}} {t}^{\mathrm{2}} }{dy} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+{y}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt}\:{dy} \\ $$$${t}=\frac{{u}}{{y}}\Rightarrow{dt}=\frac{{du}}{{y}},{t}^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} }{{y}^{\mathrm{2}} } \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{−\mathrm{5}/\mathrm{4}} \int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{y}^{\mathrm{2}} }}}{du}\:{dy} \\ $$$$\sqrt{\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{y}^{\mathrm{2}} }}=\sqrt{\frac{{y}^{\mathrm{2}} +{u}^{\mathrm{2}} }{{y}}} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{−\mathrm{1}/\mathrm{4}} \int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\sqrt{{y}^{\mathrm{2}} +{u}^{\mathrm{2}} }}{du}\:{dy} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\sqrt{{y}^{\mathrm{2}} +{u}^{\mathrm{2}} }}{dx}=\frac{\pi}{\mathrm{2}\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}}\left({y}^{−\mathrm{1}/\mathrm{4}} −{y}^{\mathrm{5}/\mathrm{4}} \right) \\ $$$${I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{y}^{−\mathrm{1}/\mathrm{4}} −{y}^{−\mathrm{5}/\mathrm{4}} }{\:\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}}{dy} \\ $$$${y}=\mathrm{sin}\theta\Rightarrow{dy}=\mathrm{cos}\:\theta\:{d}\theta \\ $$$${I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{−\mathrm{1}/\mathrm{4}} \theta−\mathrm{sin}^{\mathrm{5}/\mathrm{4}} \theta}{\:\sqrt{\mathrm{sin}\:\theta−\mathrm{1}}}\mathrm{cos}\:\theta\:{d}\theta \\ $$$$\sqrt{\mathrm{sin}^{\mathrm{2}} \theta−\mathrm{1}}={i}\:\mathrm{cos}\:\theta \\ $$$${I}=\frac{\pi}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}^{−\mathrm{1}/\mathrm{4}} \theta−\mathrm{sin}^{−\mathrm{5}/\mathrm{4}} \theta\right){d}\theta \\ $$$${I}=\frac{\pi}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}^{−\mathrm{1}/\mathrm{4}} −\mathrm{sin}^{−\mathrm{5}/\mathrm{4}} \theta\right){d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{a}} \theta\:{d}\theta=\frac{\sqrt{\pi}\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$${I}=\frac{\pi}{\mathrm{2}{i}}\left[\frac{\sqrt{\pi}\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}−\frac{\sqrt{\pi}\Gamma\left(−\frac{\mathrm{1}}{\mathrm{8}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}\right] \\ $$$$\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{\pi}{\mathrm{sin}\frac{\pi}{\mathrm{8}}},\Gamma\left(−\frac{\mathrm{1}}{\mathrm{8}}\right)=−\frac{\mathrm{8}}{\mathrm{7}}\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right) \\ $$$${I}=\frac{\pi^{\mathrm{3}/\mathrm{2}} }{\mathrm{4}{i}}\left[\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)}{\pi}\mathrm{sin}\frac{\pi}{\mathrm{8}}+\frac{\mathrm{8}\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}{\mathrm{7}\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}\right] \\ $$$$\mathrm{sin}\frac{\pi}{\mathrm{8}}=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$${I}=\frac{\pi^{\mathrm{3}/\mathrm{2}} }{\mathrm{4}{i}}\left[\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)}{\pi}\centerdot\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}+\frac{\mathrm{8}\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}{\mathrm{7}}\right] \\ $$$$\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)=\frac{\pi}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\mathrm{sin}\frac{\pi}{\mathrm{8}}} \\ $$$${I}=\frac{\pi^{\mathrm{3}/\mathrm{2}} }{\mathrm{4}{i}}\left[\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)}{\pi}\centerdot\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}+\frac{\mathrm{8}\pi}{\mathrm{7}\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\mathrm{sin}\frac{\pi}{\mathrm{8}}}\right] \\ $$$$\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{8}}},\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{8}}=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{24}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right) \\ $$$$\left(\mathrm{2}\right):{u}=\mathrm{tan}\:{x},{du}=\left(\mathrm{1}+{u}^{\mathrm{2}} \right){dx},{x}=\mathrm{arctan}\:{u} \\ $$$$\mathrm{sin}\:{x}=\frac{{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }},\frac{\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}}}{\mathrm{sin}\:{x}}=\frac{{u}^{\frac{\mathrm{1}}{\mathrm{4}}} }{\frac{{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}}={u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${dx}=\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} },{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]\Rightarrow{x}\in\left[\mathrm{0},\infty\right] \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}}}{\mathrm{sin}\:{x}}{dx}=\int_{\mathrm{0}} ^{\infty} \mathrm{arctan}\:{u}\centerdot{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{−\mathrm{1}} =\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$\mathrm{arctan}\:{u}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}}{\mathrm{1}+{v}^{\mathrm{2}} {u}^{\mathrm{2}} }{du} \\ $$$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}\centerdot{u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{v}^{\mathrm{2}} {u}^{\mathrm{2}} }{dvdu}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{4}}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{v}^{\mathrm{2}} {u}^{\mathrm{2}} }{dudv} \\ $$$${w}={u}^{\mathrm{2}} ,{du}=\frac{\mathrm{1}}{\mathrm{2}}{w}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dw} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\left({w}^{\frac{\mathrm{1}}{\mathrm{2}}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \left(\mathrm{1}+{w}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{v}^{\mathrm{2}} {w}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}{w}^{−\frac{\mathrm{1}}{\mathrm{2}}} {dw}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {w}^{\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{w}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{v}^{\mathrm{2}} {w}\right)^{−\mathrm{1}} {dw}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {w}^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}+{w}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{v}^{\mathrm{2}} {w}\right)^{−\mathrm{1}} {dw} \\ $$$${t}=\frac{{w}}{\mathrm{1}+{w}},{w}=\frac{{t}}{\mathrm{1}−{t}},{dw}=\frac{{dt}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} },{w}=\mathrm{0}\Rightarrow{t}=\mathrm{0},{w}\rightarrow\infty\Rightarrow{t}−\mathrm{1}^{−} \\ $$$$\mathrm{1}+{w}=\frac{\mathrm{1}}{\mathrm{1}−{t}},\mathrm{1}+{v}^{\mathrm{2}} {w}=\mathrm{1}+{v}^{\mathrm{2}} \frac{{t}}{\mathrm{1}−{t}}=\frac{\mathrm{1}−{t}+{v}^{\mathrm{2}} {t}}{\mathrm{1}−{t}}=\frac{\mathrm{1}+\left({v}^{\mathrm{2}} −\mathrm{1}\right){t}}{\mathrm{1}−{t}} \\ $$$${w}^{−\frac{\mathrm{7}}{\mathrm{8}}} =\left(\frac{{t}}{\mathrm{1}−{t}}\right)^{−\frac{\mathrm{7}}{\mathrm{8}}} ={t}^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{7}}{\mathrm{8}}} \\ $$$$\left(\mathrm{1}+{w}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} =\left(\frac{\mathrm{1}}{\mathrm{1}−{t}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} =\left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\mathrm{1}+{v}^{\mathrm{2}} {w}\right)^{−\mathrm{1}} =\left(\frac{\mathrm{1}+\left({v}^{\mathrm{2}} +\mathrm{1}\right){t}}{\mathrm{1}−{t}}\right)^{−\mathrm{1}} =\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+\left({v}^{\mathrm{2}} −\mathrm{1}\right){t}\right)^{−\mathrm{1}} \\ $$$${dw}=\frac{{dt}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} } \\ $$$${w}^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}+{w}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{v}^{\mathrm{2}} {w}\right)^{−\mathrm{1}} {dw}={t}^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+\left({v}^{\mathrm{2}} −\mathrm{1}\right){t}\right)^{−\mathrm{1}} \left(\mathrm{1}−{t}\right)\frac{{dt}}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }={t}^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{7}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}−\mathrm{2}} \left(\mathrm{1}+\left({v}^{\mathrm{2}} −\mathrm{1}\right){t}\right)^{−\mathrm{1}} {dt} \\ $$$$\frac{\mathrm{7}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}−\mathrm{2}=\frac{\mathrm{7}}{\mathrm{8}}+\frac{\mathrm{4}}{\mathrm{8}}+\frac{\mathrm{8}}{\mathrm{8}}−\frac{\mathrm{16}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{8}} \\ $$$${t}^{−\frac{\mathrm{7}}{\mathrm{8}}} ={t}^{−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right)} ={t}^{−\left(\frac{\mathrm{8}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}\right)} ={t}^{−\frac{\mathrm{8}}{\mathrm{8}}+\mathrm{1}} ={t}^{−\left(\frac{{b}−\mathrm{1}}{{b}}\right)} \\ $$$${b}−\mathrm{1}=\frac{\mathrm{7}}{\mathrm{8}},{b}=\mathrm{1}+\frac{\mathrm{7}}{\mathrm{8}}=\frac{\mathrm{15}}{\mathrm{8}} \\ $$$${b}=\frac{\mathrm{5}}{\mathrm{8}},{c}−{b}=\frac{\mathrm{7}}{\mathrm{8}},{c}={b}+\frac{\mathrm{7}}{\mathrm{8}}=\frac{\mathrm{5}}{\mathrm{8}}+\frac{\mathrm{7}}{\mathrm{8}}=\frac{\mathrm{12}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{2}},{a}=\mathrm{1},{z}=−\left({v}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{1}−{v}^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{b}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\left({c}−{b}\right)−\mathrm{1}} \left(\mathrm{1}−{zt}\right)^{−{a}} {dt}={B}\left({b},{c}−{b}\right)\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b};{c};{z}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{5}}{\mathrm{8}}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{7}}{\mathrm{8}}−\mathrm{1}} \left(\mathrm{1}+\left({v}^{\mathrm{2}} +\mathrm{1}\right){t}\right)^{−\mathrm{1}} {dt}={B}\left(\frac{\mathrm{5}}{\mathrm{8}},\frac{\mathrm{7}}{\mathrm{8}}\right){F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{5}}{\mathrm{8}};\frac{\mathrm{3}}{\mathrm{2}};\mathrm{1}−{v}^{\mathrm{2}} \right) \\ $$$${B}\left(\frac{\mathrm{5}}{\mathrm{8}},\frac{\mathrm{7}}{\mathrm{8}}\right)=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)},\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {w}^{−\frac{\mathrm{7}}{\mathrm{8}}} \left(\mathrm{1}+{w}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{v}^{\mathrm{2}} {w}\right)^{−\mathrm{1}} {dw}=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}{\frac{\sqrt{\pi}}{\mathrm{2}}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{5}}{\mathrm{8}};\frac{\mathrm{3}}{\mathrm{2}};\mathrm{1}−{v}^{\mathrm{2}} \right)=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}{\:\sqrt{\pi}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{5}}{\mathrm{8}};\frac{\mathrm{3}}{\mathrm{2}};\mathrm{1}−{v}^{\mathrm{2}} \right) \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}{\:\sqrt{\pi}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{5}}{\mathrm{8}};\frac{\mathrm{3}}{\mathrm{2}};\mathrm{1}−{v}^{\mathrm{2}} \right){dv} \\ $$$$\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{5}}{\mathrm{8}};\frac{\mathrm{3}}{\mathrm{2}};\mathrm{1}−{v}^{\mathrm{2}} \right){dv}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}\right)_{{n}} \left(\frac{\mathrm{5}}{\mathrm{8}}\right)_{{n}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} {n}!}\left(\mathrm{1}−{v}^{\mathrm{2}} \right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{5}}{\mathrm{8}};\frac{\mathrm{3}}{\mathrm{2}};\mathrm{1}−{v}^{\mathrm{2}} \right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}\right)_{{n}} \left(\frac{\mathrm{5}}{\mathrm{8}}\right)_{{n}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} {n}!}\left(\mathrm{1}−{v}^{\mathrm{2}} \right)^{{n}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{5}}{\mathrm{8}};\frac{\mathrm{3}}{\mathrm{2}};\mathrm{1}−{v}^{\mathrm{2}} \right)=\int_{\mathrm{0}\:} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{5}}{\mathrm{8}}\right)_{{n}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} {n}!}\left(\mathrm{1}−{v}^{\mathrm{2}} \right)^{{n}} {dv} \\ $$$${s}={v}^{\mathrm{2}} ,{dv}=\frac{{ds}}{\mathrm{2}{s}^{\frac{\mathrm{1}}{\mathrm{2}}} },{v}:\mathrm{0}\rightarrow\mathrm{1}\Rightarrow{s}:\mathrm{0}\rightarrow\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{s}\right)^{{n}} \frac{{ds}}{\mathrm{2}{s}^{\frac{\mathrm{1}}{\mathrm{2}}} }=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {s}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{s}\right)^{{n}} {ds}=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{1}}{\mathrm{2}},{n}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi},\Gamma\left({n}+\mathrm{1}\right)={n}! \\ $$$$\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left(\left({n}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}+\mathrm{1}} }\sqrt{\pi}=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}+\mathrm{1}} {n}!}\sqrt{\pi} \\ $$$${B}\left(\frac{\mathrm{1}}{\mathrm{2}},{n}+\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {s}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}\left(\mathrm{1}−{s}\right)^{\left({n}−\mathrm{1}\right)−\mathrm{1}} {ds}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right){n}+\mathrm{1}}{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}=\frac{\sqrt{\pi}{n}!}{\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} {n}!}\sqrt{\pi}}=\frac{\left({n}!\right)^{\mathrm{2}} \mathrm{4}^{{n}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}\right)}} \\ $$$$\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)}{\mathrm{2}^{{n}+\mathrm{1}} }=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!!}{\mathrm{2}^{{n}+\mathrm{1}} }=\sqrt{\pi}=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}+\mathrm{1}} {n}!}\sqrt{\pi} \\ $$$${B}\left(\frac{\mathrm{1}}{\mathrm{2}},{n}+\mathrm{1}\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}=\frac{\sqrt{\pi}{n}}{\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} {n}!}\sqrt{\pi}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{1}}{\mathrm{2}},{n}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\left({n}!\right)^{\mathrm{2}} \mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\frac{\left({n}!\right)^{\mathrm{2}} \mathrm{2}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)!=\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)! \\ $$$$\frac{\left({n}!\right)^{\mathrm{2}} \mathrm{2}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\frac{\left({n}!\right)^{\mathrm{2}} \mathrm{4}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{1}}{\mathrm{2}},{n}+\mathrm{1}\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\mathrm{1}\right)}{\mathrm{2}\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}=\frac{\sqrt{\pi}{n}!}{\mathrm{2}\centerdot\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}{\mathrm{2}^{{n}+\mathrm{1}} }\sqrt{\pi}}=\frac{\left({n}!\right)^{\mathrm{2}} \mathrm{4}^{{n}} }{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\left(\mathrm{2}{n}+\mathrm{1}\right){n}!} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{v}^{\mathrm{2}} \right)^{{n}} {dv}=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{1}}{\mathrm{2}},{n}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\sqrt{\pi}{n}!}{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)} \\ $$$$\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}{\mathrm{2}^{{n}+\mathrm{1}} }\sqrt{\pi}=\frac{\left(\mathrm{2}{n}−\mathrm{1}\right)!}{\mathrm{4}^{{n}} \left({n}!\right)^{\mathrm{2}} \mathrm{2}^{{n}+\mathrm{1}} }\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\pi} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\sqrt{\pi}{n}!}{\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}{\mathrm{2}^{{n}+\mathrm{1}} }\sqrt{\pi}}=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{n}!\mathrm{2}^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}=\frac{\mathrm{2}^{{n}} {n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)!!} \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)!!=\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} {n}!} \\ $$$$\frac{\mathrm{2}^{{n}} {n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)!!}=\frac{\mathrm{2}^{{n}} {n}!}{\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\mathrm{2}^{{n}} {n}!}}=\frac{\mathrm{4}^{{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{5}}{\mathrm{8}};\frac{\mathrm{3}}{\mathrm{2}};\mathrm{1}−{v}^{\mathrm{2}} \right){dv}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{5}}{\mathrm{8}}\right)_{{n}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} {n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{v}^{\mathrm{2}} \right)^{{n}} {dv}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{5}}{\mathrm{8}}\right)_{{n}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} {n}!}\centerdot\frac{\mathrm{4}^{{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{5}}{\mathrm{8}}\right)_{{n}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} {n}!}\centerdot\frac{\mathrm{4}^{{n}} {n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\centerdot\frac{\mathrm{1}}{{n}!} \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)!=\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}!\right),\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}=\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} },\frac{\mathrm{4}^{{n}} {n}!}{\left(\mathrm{2}{n}\right)!}=\frac{\mathrm{1}}{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}} \\ $$$$\frac{\mathrm{4}^{{n}} {n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\frac{\mathrm{4}^{{n}} {n}!}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}\right)!}=\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{5}}{\mathrm{8}}\right)_{{n}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} {n}!}\centerdot\frac{\mathrm{4}^{{n}} \left({n}!\right)^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{5}}{\mathrm{8}}\right)_{{n}} \left(\mathrm{1}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} \left(\mathrm{1}\right)_{{n}} {n}!}\centerdot\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}} \\ $$$${I}=\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{5}}{\mathrm{8}}\right)_{{n}} \left(\mathrm{1}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} \left(\mathrm{1}\right)_{{n}} {n}!}\centerdot\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}} \\ $$$$\Gamma\left(\frac{\mathrm{5}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)=\Gamma\left(\mathrm{1}−\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{\pi}{\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}\centerdot\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)} \\ $$$$\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)=\mathrm{sin}\left(\pi−\frac{\mathrm{5}\pi}{\mathrm{8}}\right)=\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right),\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)=\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right) \\ $$$$\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Gamma\left(\frac{\mathrm{5}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)}=\frac{\pi^{\mathrm{2}} }{\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}}=\frac{\mathrm{4}\pi^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}}=\mathrm{2}\sqrt{\mathrm{2}}\pi^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}}}{\mathrm{sin}\:{x}}{dx}=\frac{\left(\mathrm{3}−\sqrt{\mathrm{2}}\right)\sqrt{\pi}}{\mathrm{24}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right) \\ $$

Question Number 221544    Answers: 1   Comments: 0

((x−1))^(1/(x+1)) = ((x+1))^(1/(x−1)) , x real

$$\:\:\sqrt[{{x}+\mathrm{1}}]{{x}−\mathrm{1}}\:=\:\sqrt[{{x}−\mathrm{1}}]{{x}+\mathrm{1}}\:,\:{x}\:{real}\: \\ $$

Question Number 221527    Answers: 0   Comments: 9

(1):∫_0 ^1 K(k)^3 dk=(1/8)∫∫∫_([0,1]^3 ) u_1 ^(−(1/2)) u_2 ^(−(1/2)) u_3 ^(−(1/2)) (1−u_1 )^(−(1/2)) (1−u_2 )^(−(1/2)) (1−u_3 )^(−(( 1)/2)) (∫_0 ^1 Π_(i=1) ^3 (1−k^2 u_i )^(−(1/2)) dk)du_1 du_2 du_3 =(1/(16))Σ_(n_1 =0) ^∞ Σ_(n_2 =0) ^∞ Σ_(n_3 =0) ^∞ (((2n_1 )),(n_1 ) ) (((2n_2 )),(n_2 ) ) (((2n_3 )),(n_3 ) )(1/(4^(n_1 +n_2 +n_3 ) (n_1 +n_2 +n_3 +(1/2))))∫∫∫_([0,1]^3 ) Π_(i=1) ^3 u_i ^(n_i −(1/2)) (1−u_i )^(−(1/2)) du_1 du_2 du_3 =(1/(16))Σ_(n_1 =0) ^∞ Σ_(n_2 =0) ^∞ Σ_(n_3 =0) ^∞ (((2n_1 )),(n_1 ) ) (((2n_2 )),(n_2 ) ) (((2n_3 )),(n_3 ) )(1/(4^(n_1 +n_2 +n_3 ) (n_1 +n_2 +n_3 +(1/2))))Π_(i=1) ^3 ∫_0 ^1 u_i ^(n_i −(1/2)) (1−u_i )^(−(1/2)) du_i =(1/(16))Σ_(n_1 =0) ^∞ Σ_(n_2 =0) ^∞ Σ_(n_3 =0) ^∞ (((2n_1 )),(n_1 ) ) (((2n_2 )),(n_2 ) ) (((2n_3 )),(n_3 ) )(1/(4^(n_1 +n_2 +n_3 ) (n_1 +n_2 +n_3 +(1/2))))Π_(i=1) ^3 (((2n_i )),(n_i ) )(π/4^n_i ) =(1/(16))π^3 Σ_(n_1 =0) ^∞ Σ_(n_2 =0) ^∞ Σ_(n_3 =0) ^∞ ((Π_(i=1) ^3 (((2n_i )),(n_i ) )^2 )/(16^(n_1 +n_2 +n_3 ) (n_(1 ) +n_2 +n_3 +(1/2)))) =((3Γ((1/4))^8 )/(1280π^3 )) (2):K(k)=(π/2)Σ_(n=0) ^∞ ((((2n)!)/(2^(2n) (n!)^2 )))^2 k^(2n) K(k)^3 =((π/2))^3 Σ_(n,m,p=0) ^∞ ((((2n)!(2m)!(2p)!)/(2^(2(n+m+p)) (n!m!p!)^2 )))^2 k^(2(n+m+p)) ∫_0 ^1 K(k)^3 dk=((π/2))^3 Σ_(n,m,p=0) ^∞ ((((2n)!(2m)!(2p)!)/(2^(2(n+m+p)) (n!m!p!)^2 )))^2 (1/(2(n+m+p)+1)) Σ_(n,m,p=0) ^∞ (((2n)!(2m)!(2p)!)/((n!m!p!)^2 )) (x^(n+m+p) /(2(n+m+p)+1))=(3/5) _4 F_3 ((1/2),(1/2),(1/2),(1/2);1,1,1;16x) ∵x=(1/(16))⇒ _4 F_3 ((1/2),(1/2),(1/2),(1/2);1,1,1;1)=((Γ((1/4))^8 )/(128π^6 )) ∴∫_0 ^1 K(k)^3 dk=((3π^3 )/(1280))∙((Γ((1/4))^8 )/π^6 )=((3Γ((1/4))^8 )/(1280π^3 )) (1)∫_0 ^1 k^(1/3) (1−k^2 )^(−1/3) K(k)^2 dk k=sinθ⇒dk=cosθdθ ∫_(0 ) ^(π/2) sin^(1/3) θ(1−sin^2 θ)^(−1/3) cosθK(sin θ)^2 dθ =∫_0 ^(π/2) sin^(1/3) θ cos^(−2/3) θ K(sin θ)^2 dθ K(k)=(π/2)Σ_(n=0) ^∞ ((((1/2))_n ^2 )/((n!)^2 ))k^(2n) K(k)^2 =((π/2))^2 Σ_(m,n=0) ^∞ ((((1/2))_m ^2 ((1/2))_n ^2 )/((m!)(n!)^2 ))k^(2(m+n)) ∫_0 ^(π/2) sin^(1/3+2(m+n)) θ cos^(−2/3) θ dθ=(1/2)B((2/3)+m+n,(1/6)) =(1/2) ((Γ((2/3)+m+n)Γ((1/6)))/(Γ((5/6)+m+n))) Σ_(m,n=0) ^∞ ((((1/2))_m ^2 ((1/2))_n ^2 )/((m!)^2 (n!)^2 )) ((Γ((2/3)+m+n)Γ((1/6)))/(2Γ((5/6)+m+n))) =((Γ((1/6)))/2)Σ_(k=0) ^∞ ((Γ((2/3)+k))/(Γ((5/6)+k)))Σ_(m+n=k) ((((1/2))_m ^2 ((1/2))_n ^2 )/((m!)^2 (n!)^2 )) =((Γ((1/6)))/2)Σ_(k=0) ^∞ ((Γ((2/3)+k))/(Γ((5/6)+k))) ((((1/2))_k ^2 )/((k!)^2 )) (((2k)),(k) ) =((Γ((1/6)))/2)Σ_(k=0) ^∞ ((Γ((2/3)+k)Γ((1/2)+k)((1/2))_k )/(Γ((5/6)+k)Γ(1+k))) ((((1/2))_k )/(k!)) =((Γ((1/6)))/2) ((Γ((2/3))Γ((1/2)))/(Γ((5/6))))F_2 ((2/3),(1/2),(1/2);(5/6),1;1) =((Γ((1/6))Γ((2/3))Γ((1/2)))/(2Γ((5/6)))) ((Γ((5/6))Γ((1/6)))/(Γ((1/2))Γ((2/3))))=((Γ((1/6))^2 )/2) (π^2 /4) ((Γ((1/6))^2 )/2)=((π^2 Γ((1/6))^2 )/8) “(((√3)Γ((1/3))^9 )/(64(2)^(1/3) π^3 ))”=((π^2 Γ((1/6))^2 )/8)⇒Γ((1/6))=(((√3)Γ((1/3))^(9/2) )/(8(2)^(1/6) π^(5/2) )) (2):Γ((2/3))≜((2π)/( (√3)Γ((1/3)))) K(k)=(π/2) _2 F_1 ((1/2),(1/2);1;k^2 ) K(k)^2 =((π/2))^2 Σ_(m=0) ^∞ Σ_(n=0) ^∞ ((((1/2))_m ((1/2))_n ((1/2))_m ((1/2))_n )/(m!n!m!n!))k^(2(m+n)) ((1/2))_m = (((2m)),(m) )(1/4^m ) ∀m∈N K(k)^2 =(π^4 /4)Σ_(p=0) ^∞ (((2p)),(p) )(k^(2p) /(16^p ))Σ_(q=0) ^p (((2q)),(q) )^2 (((2(p−q))),((p−q)) )( (((2p)),(p) )^2 /( (((2q)),(q) )^2 (((2(p−q))),((p−q)) )^2 )) c_p ≜Σ_(p=0) ^p (((2q)),(q) )^2 (((2(p−q))),((p−q)) )^2 K(k)^2 =(π^2 /4)Σ_(p=0) ^∞ c_p (k^(2p) /(16^p )) ∫_0 ^1 k^(1/3) (1−k^2 )^(−(1/3)) K(k)^2 dk=(π^2 /4)Σ_(p=0) ^∞ c_p (1/(16^p ))∫_0 ^1 k^(1/3) (1−k^2 )^(−(1/3)) k^(2p) dk u≜k^2 ⇒dk=(1/2)u^(−(1/2)) du ∫_0 ^1 k^((1/3)+2p) (1−k^2 )^(−(1/3)) dk=(1/2)∫_0 ^1 u^((1/6)+p) (1−u)^(−(1/3)) u^(−(1/2)) du=(1/2)∫_0 ^1 u^(p−(1/3)) (1−u)^(−(1/3)) du ∫_0 ^1 u^(a−1) (1−u)^(b−1) du=B(a,b) for a>0,b a≜p+(2/3),b≜(2/3) B(p+(2/3),(2/3))=∫_0 ^1 u^(p+(2/3)−1) (1−u)^((2/3)−1) du Γ(p+(3/4))=(p+(2/3))Γ(p+(2/3)) B(p+(2/3),(2/3))=((Γ(p+(2/3))Γ((2/3)))/(Γ(p+(3/4))))=((Γ(p+(2/3))Γ((2/3)))/((p+(2/3))(p+(2/3))))=((Γ((2/3)))/(p+(2/3))) ∫_0 ^1 k^((1/3)+2p) (1−k^2 )^(−(1/3)) dk=(1/2) ((Γ((2/3)))/(p+(2/3))) ∫_0 ^1 k^(1/3) (1−k^2 )^(−(1/2)) K(k)^2 dk=(π^2 /4)Σ_(p=0) ^∞ c_p (1/(16^p )) (1/2) ((Γ((2/3)))/(p+(2/3)))=(π^2 /8)Γ((2/3))Σ_(p=0) ^∞ c_p (1/(16^p )) (1/(p+(2/3))) (1/(p+(2/3)))=∫_0 ^1 v^(p+(2/3)−1) dv ∀p∈N Σ_(p=0) ^∞ c_p ((1/(16)))^p (1/(p+(2/3)))=∫_0 ^1 v^((2/3)−1) Σ_(p=0) ^∞ c_p ((v/(16)))^p dv Σ_(p=0) ^∞ c_p w^p =(Σ_(p=0) ^∞ (((2n)),(n) )^2 w^n )=( _2 F_1 ((1/2),(1/2);1;16w))^2 Σ_(p=0) ^∞ c_(p ) ((1/(16)))^p +(1/(p+(2/3)))=∫_0 ^1 v^((2/3)−1) ( _2 F_1 ((1/2);(1/2);1;16∙(v/(16))))^2 dv=∫_0 ^1 v^((2/3)−1) ( _2 F_1 ((1/2),(1/2);1;v))^2 dv _2 F_1 ((1/2),(1/2);1,v)=(2/π)K((√v))^2 Σ_(p=0) ^∞ c_p ((1/(16)))^p (1/(p+(2/3)))=∫_0 ^1 v^((2/3)−1) (4/π^2 )K((√v))^2 dv t≜(√v)⇒v=t^2 ⇒dv=2tdt ∫_0 ^1 v^((2/3)−1) K((√v))^2 dv=∫_0 ^1 (t^2 )^((2/3)−1) K(t)^2 2tdt=2∫_0 ^1 t^((4/3)−2) −K(t)^2 tdt=2∫_0 ^1 t^(1/3) K(t)^2 dt Σ_(p=0) ^∞ c_p ((1/(16)))^p (1/(p+(2/3)))=(4/π^2 )∙2∫_0 ^1 t^(1/3) K(t)^2 dt=(8/π^2 )∫_0 ^1 t^(1/3) K(t)^2 dt ∫_0 ^1 k^(1/3) (1−k^2 )^(−(1/3)) K(k)^2 dk=(π^2 /8)Γ((3/8))∙(8/π^2 )∫_0 ^1 t^(1/3) K(t)^2 dt=Γ((2/3))∫_0 ^1 t^(1/3) K(t)^2 dt ∫_0 ^1 t^(1/3) K(t)^2 dt=∫_0 ^1 t^(1/3) (1−t^2 )^(−(1/3)) K(t)^2 dt (1−t^2 )^(−(1/3)) =Γ((3/4))((Γ((2/3)))/(Γ((2/3))))(1−t^2 )^(−(1/3)) ∫_(0 ) ^1 t^(1/3) (1−t^2 )^(−(1/3)) K(t)^2 dt=Γ((2/3))∫_0 ^1 t^(1/3) K(t)^2 dt Γ((2/3))∫_0 ^1 t^(1/3) K(t)^2 dt=Γ((2/3))∙((Γ((2/3)))/(Γ((2/3))))∫_0 ^1 t^(1/3) K(t)^2 dt ∫_0 ^∞ (t^(α−1) /(t^π +1))=(1/(sin α)) For 0<α<π Γ((1/3))^3 =(2/π)Γ((1/3))^3 π=2(√3)Γ((1/3))^2 Γ((1/3))(π/(2Γ((1/3)))) Γ((1/3))Γ((2/3))=((2π)/( (√3))) Γ((1/9))^9 =Γ((1/3))^9 (((√3)Γ((1/3))^9 )/(64(2)^(1/3) π^3 ))=((Γ((1/9))^9 Γ((1/3))Γ((2/3))sin((π/3)))/(64(2)^(1/3) π^3 )) ((Γ((1/3))Γ((2/3)))/(Γ((1/3))Γ((2/3))))π ((Γ((1/3))^9 Γ((1/3))Γ((2/3))((√3)/2))/(64(2)^(1/3) π^3 π)) ((Γ((1/3))Γ((2/3)))/(Γ((1/3))Γ((2/3)))) ((Γ((1/3))^9 Γ((1/3))Γ((2/3)))/(128(2)^(1/8) π^4 Γ((1/3))Γ((2/3)))) ((√3)/2)=((Γ((1/3))^9 (√3))/(256(2)^(1/3) π^4 )) =(((√3)Γ((1/3))^9 )/(64(2)^(1/3) π^3 )) ∫_0 ^(π/2) ((x((tan x))^(1/4) )/(sin x))dx=(((3−2(√2)))/(24))Γ((1/8))Γ((3/8)) (1):I=∫_0 ^(π/2) ((x tan^(1/4) x)/(sin x))dx=∫_0 ^(π/2) x tan^(−3/4) x sec x dx tan x=t⇒dx=(dt/(1+t^2 )),sec x=(√(1+t^2 )) I=∫_0 ^∞ ((arctant∙t^(−3/4) )/( (√(1+t^2 ))))dt arctan t=∫_0 ^1 (t/(1+y^2 t^2 ))dy I=∫_0 ^1 ∫_0 ^∞ (t^(1/4) /((1+y^2 t^2 )(√(1+t^2 ))))dt dy t=(u/y)⇒dt=(du/y),t^2 =(u^2 /y^2 ) I=∫_0 ^1 y^(−5/4) ∫_0 ^∞ (u^(1/4) /((1+u^2 )(√(1+(u^2 /y^2 )))))du dy (√(1+(u^2 /y^2 )))=(√((y^2 +u^2 )/y)) I=∫_0 ^1 y^(−1/4) ∫_0 ^∞ (u^(1/4) /((1+u^2 )(√(y^2 +u^2 ))))du dy ∫_0 ^∞ (u^(1/4) /((1+u^2 )(√(y^2 +u^2 ))))dx=(π/(2(√(y^2 −1))))(y^(−1/4) −y^(5/4) ) I=(π/2)∫_0 ^1 ((y^(−1/4) −y^(−5/4) )/( (√(y^2 −1))))dy y=sinθ⇒dy=cos θ dθ I=(π/2)∫_0 ^(π/2) ((sin^(−1/4) θ−sin^(5/4) θ)/( (√(sin θ−1))))cos θ dθ (√(sin^2 θ−1))=i cos θ I=(π/(2i))∫_0 ^(π/2) (sin^(−1/4) θ−sin^(−5/4) θ)dθ I=(π/(2i))∫_0 ^(π/2) (sin^(−1/4) −sin^(−5/4) θ)dθ ∫_0 ^(π/2) sin^a θ dθ=(((√π)Γ(((a+1)/2)))/(2Γ((a/2)+1))) I=(π/(2i))[(((√π)Γ((3/8)))/(2Γ((7/8))))−(((√π)Γ(−(1/8)))/(2Γ((3/8))))] Γ((7/8))Γ((1/8))=(π/(sin(π/8))),Γ(−(1/8))=−(8/7)Γ((7/8)) I=(π^(3/2) /(4i))[((Γ((3/8))Γ((1/8)))/π)sin(π/8)+((8Γ((7/8))Γ((3/8)))/(7Γ((3/8))))] sin(π/8)=((√(2−(√2)))/2) I=(π^(3/2) /(4i))[((Γ((3/8))Γ((1/8)))/π)∙((√(2−(√2)))/2)+((8Γ((7/8)))/7)] Γ((7/8))=(π/(Γ((1/8))sin(π/8))) I=(π^(3/2) /(4i))[((Γ((3/8))Γ((1/8)))/π)∙((√(2−(√2)))/2)+((8π)/(7Γ((1/8))sin(π/8)))] Γ((3/8))Γ((5/4))=(π/(sin((3π)/8))),sin((3π)/8)=((√(2+(√2)))/2) =(((3−2(√2)))/(24))Γ((1/8))Γ((3/8))

$$\left(\mathrm{1}\right):\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{\mathrm{K}}\left({k}\right)^{\mathrm{3}} {dk}=\frac{\mathrm{1}}{\mathrm{8}}\int\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } {u}_{\mathrm{1}} ^{−\frac{\mathrm{1}}{\mathrm{2}}} {u}_{\mathrm{2}} ^{−\frac{\mathrm{1}}{\mathrm{2}}} {u}_{\mathrm{3}} ^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}_{\mathrm{1}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}_{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}_{\mathrm{3}} \right)^{−\frac{\:\mathrm{1}}{\mathrm{2}}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\left(\mathrm{1}−{k}^{\mathrm{2}} {u}_{{i}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dk}\right){du}_{\mathrm{1}} {du}_{\mathrm{2}} {du}_{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\underset{{n}_{\mathrm{1}} =\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}_{\mathrm{2}} =\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}_{\mathrm{3}} =\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}_{\mathrm{1}} }\\{{n}_{\mathrm{1}} }\end{pmatrix}\begin{pmatrix}{\mathrm{2}{n}_{\mathrm{2}} }\\{{n}_{\mathrm{2}} }\end{pmatrix}\begin{pmatrix}{\mathrm{2}{n}_{\mathrm{3}} }\\{{n}_{\mathrm{3}} }\end{pmatrix}\frac{\mathrm{1}}{\mathrm{4}^{{n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} } \left({n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\right)}\int\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } \underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}{u}_{{i}} ^{{n}_{{i}} −\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}_{{i}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du}_{\mathrm{1}} {du}_{\mathrm{2}} {du}_{\mathrm{3}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\underset{{n}_{\mathrm{1}} =\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}_{\mathrm{2}} =\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}_{\mathrm{3}} =\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}_{\mathrm{1}} }\\{{n}_{\mathrm{1}} }\end{pmatrix}\begin{pmatrix}{\mathrm{2}{n}_{\mathrm{2}} }\\{{n}_{\mathrm{2}} }\end{pmatrix}\begin{pmatrix}{\mathrm{2}{n}_{\mathrm{3}} }\\{{n}_{\mathrm{3}} }\end{pmatrix}\frac{\mathrm{1}}{\mathrm{4}^{{n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} } \left({n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\right)}\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}_{{i}} ^{{n}_{{i}} −\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}_{{i}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du}_{{i}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\underset{{n}_{\mathrm{1}} =\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}_{\mathrm{2}} =\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}_{\mathrm{3}} =\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}_{\mathrm{1}} }\\{{n}_{\mathrm{1}} }\end{pmatrix}\begin{pmatrix}{\mathrm{2}{n}_{\mathrm{2}} }\\{{n}_{\mathrm{2}} }\end{pmatrix}\begin{pmatrix}{\mathrm{2}{n}_{\mathrm{3}} }\\{{n}_{\mathrm{3}} }\end{pmatrix}\frac{\mathrm{1}}{\mathrm{4}^{{n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} } \left({n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\right)}\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\begin{pmatrix}{\mathrm{2}{n}_{{i}} }\\{{n}_{{i}} }\end{pmatrix}\frac{\pi}{\mathrm{4}^{{n}_{{i}} } } \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\pi^{\mathrm{3}} \underset{{n}_{\mathrm{1}} =\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}_{\mathrm{2}} =\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}_{\mathrm{3}} =\mathrm{0}} {\overset{\infty} {\sum}}\frac{\underset{{i}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\begin{pmatrix}{\mathrm{2}{n}_{{i}} }\\{{n}_{{i}} }\end{pmatrix}^{\mathrm{2}} }{\mathrm{16}^{{n}_{\mathrm{1}} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} } \left({n}_{\mathrm{1}\:} +{n}_{\mathrm{2}} +{n}_{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{3}\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{8}} }{\mathrm{1280}\pi^{\mathrm{3}} } \\ $$$$\left(\mathrm{2}\right):\boldsymbol{{K}}\left({k}\right)=\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(\mathrm{2}{n}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left({n}!\right)^{\mathrm{2}} }\right)^{\mathrm{2}} {k}^{\mathrm{2}{n}} \\ $$$$\boldsymbol{\mathrm{K}}\left({k}\right)^{\mathrm{3}} =\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{3}} \underset{{n},{m},{p}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(\mathrm{2}{n}\right)!\left(\mathrm{2}{m}\right)!\left(\mathrm{2}{p}\right)!}{\mathrm{2}^{\mathrm{2}\left({n}+{m}+{p}\right)} \left({n}!{m}!{p}!\right)^{\mathrm{2}} }\right)^{\mathrm{2}} {k}^{\mathrm{2}\left({n}+{m}+{p}\right)} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{\mathrm{K}}\left({k}\right)^{\mathrm{3}} {dk}=\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{3}} \underset{{n},{m},{p}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\left(\mathrm{2}{n}\right)!\left(\mathrm{2}{m}\right)!\left(\mathrm{2}{p}\right)!}{\mathrm{2}^{\mathrm{2}\left({n}+{m}+{p}\right)} \left({n}!{m}!{p}!\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}\left({n}+{m}+{p}\right)+\mathrm{1}} \\ $$$$\underset{{n},{m},{p}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{n}\right)!\left(\mathrm{2}{m}\right)!\left(\mathrm{2}{p}\right)!}{\left({n}!{m}!{p}!\right)^{\mathrm{2}} }\:\frac{{x}^{{n}+{m}+{p}} }{\mathrm{2}\left({n}+{m}+{p}\right)+\mathrm{1}}=\frac{\mathrm{3}}{\mathrm{5}}\:_{\mathrm{4}} {F}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1},\mathrm{1},\mathrm{1};\mathrm{16}{x}\right) \\ $$$$\because{x}=\frac{\mathrm{1}}{\mathrm{16}}\Rightarrow\:_{\mathrm{4}} {F}_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1},\mathrm{1},\mathrm{1};\mathrm{1}\right)=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{8}} }{\mathrm{128}\pi^{\mathrm{6}} } \\ $$$$\therefore\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{K}}\left({k}\right)^{\mathrm{3}} {dk}=\frac{\mathrm{3}\pi^{\mathrm{3}} }{\mathrm{1280}}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{8}} }{\pi^{\mathrm{6}} }=\frac{\mathrm{3}\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{8}} }{\mathrm{1280}\pi^{\mathrm{3}} } \\ $$$$\left(\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {k}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{−\mathrm{1}/\mathrm{3}} \boldsymbol{{K}}\left({k}\right)^{\mathrm{2}} {dk} \\ $$$${k}=\mathrm{sin}\theta\Rightarrow{dk}=\mathrm{cos}\theta{d}\theta \\ $$$$\int_{\mathrm{0}\:} ^{\pi/\mathrm{2}} \mathrm{sin}^{\mathrm{1}/\mathrm{3}} \theta\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta\right)^{−\mathrm{1}/\mathrm{3}} \mathrm{cos}\theta\boldsymbol{{K}}\left(\mathrm{sin}\:\theta\right)^{\mathrm{2}} {d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}^{\mathrm{1}/\mathrm{3}} \theta\:\mathrm{cos}^{−\mathrm{2}/\mathrm{3}} \theta\:\boldsymbol{{K}}\left(\mathrm{sin}\:\theta\right)^{\mathrm{2}} {d}\theta \\ $$$$\boldsymbol{{K}}\left({k}\right)=\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }{\left({n}!\right)^{\mathrm{2}} }{k}^{\mathrm{2}{n}} \\ $$$$\boldsymbol{{K}}\left({k}\right)^{\mathrm{2}} =\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \underset{{m},{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }{\left({m}!\right)\left({n}!\right)^{\mathrm{2}} }{k}^{\mathrm{2}\left({m}+{n}\right)} \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{sin}^{\mathrm{1}/\mathrm{3}+\mathrm{2}\left({m}+{n}\right)} \theta\:\mathrm{cos}^{−\mathrm{2}/\mathrm{3}} \theta\:{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{2}}{\mathrm{3}}+{m}+{n},\frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}+{m}+{n}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}+{m}+{n}\right)} \\ $$$$\underset{{m},{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} ^{\mathrm{2}} }{\left({m}!\right)^{\mathrm{2}} \left({n}!\right)^{\mathrm{2}} }\:\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}+{m}+{n}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}+{m}+{n}\right)} \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}+{k}\right)}{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}+{k}\right)}\underset{{m}+{n}={k}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\overset{\mathrm{2}} {\right)}_{{n}} }{\left({m}!\right)^{\mathrm{2}} \left({n}!\right)^{\mathrm{2}} } \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}+{k}\right)}{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}+{k}\right)}\:\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} ^{\mathrm{2}} }{\left({k}!\right)^{\mathrm{2}} }\begin{pmatrix}{\mathrm{2}{k}}\\{{k}}\end{pmatrix} \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{2}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}+{k}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}+{k}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}+{k}\right)\Gamma\left(\mathrm{1}+{k}\right)}\:\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{k}} }{{k}!} \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\mathrm{2}}\:\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}{F}_{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{5}}{\mathrm{6}},\mathrm{1};\mathrm{1}\right) \\ $$$$=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)}\:\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{6}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\pi^{\mathrm{2}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\underline{``\frac{\sqrt{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} }{\mathrm{64}\sqrt[{\mathrm{3}}]{\mathrm{2}}\pi^{\mathrm{3}} }''}=\frac{\pi^{\mathrm{2}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} }{\mathrm{8}}\Rightarrow\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)=\frac{\sqrt{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}/\mathrm{2}} }{\mathrm{8}\sqrt[{\mathrm{6}}]{\mathrm{2}}\pi^{\mathrm{5}/\mathrm{2}} } \\ $$$$\left(\mathrm{2}\right):\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\triangleq\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$\boldsymbol{\mathrm{K}}\left({k}\right)=\frac{\pi}{\mathrm{2}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};{k}^{\mathrm{2}} \right) \\ $$$$\boldsymbol{{K}}\left({k}\right)^{\mathrm{2}} =\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} \underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{m}!{n}!{m}!{n}!}{k}^{\mathrm{2}\left({m}+{n}\right)} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{m}} =\begin{pmatrix}{\mathrm{2}{m}}\\{{m}}\end{pmatrix}\frac{\mathrm{1}}{\mathrm{4}^{{m}} }\:\forall{m}\in\mathbb{N} \\ $$$$\boldsymbol{{K}}\left({k}\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{4}} }{\mathrm{4}}\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{p}}\\{{p}}\end{pmatrix}\frac{{k}^{\mathrm{2}{p}} }{\mathrm{16}^{{p}} }\underset{{q}=\mathrm{0}} {\overset{{p}} {\sum}}\begin{pmatrix}{\mathrm{2}{q}}\\{{q}}\end{pmatrix}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}\left({p}−{q}\right)}\\{{p}−{q}}\end{pmatrix}\frac{\begin{pmatrix}{\mathrm{2}{p}}\\{{p}}\end{pmatrix}^{\mathrm{2}} }{\begin{pmatrix}{\mathrm{2}{q}}\\{{q}}\end{pmatrix}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}\left({p}−{q}\right)}\\{{p}−{q}}\end{pmatrix}^{\mathrm{2}} } \\ $$$${c}_{{p}} \triangleq\underset{{p}=\mathrm{0}} {\overset{{p}} {\sum}}\begin{pmatrix}{\mathrm{2}{q}}\\{{q}}\end{pmatrix}^{\mathrm{2}} \begin{pmatrix}{\mathrm{2}\left({p}−{q}\right)}\\{{p}−{q}}\end{pmatrix}^{\mathrm{2}} \\ $$$$\boldsymbol{{K}}\left({k}\right)^{\mathrm{2}} =\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}} \frac{{k}^{\mathrm{2}{p}} }{\mathrm{16}^{{p}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {k}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({k}\right)^{\mathrm{2}} {dk}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}} \frac{\mathrm{1}}{\mathrm{16}^{{p}} }\int_{\mathrm{0}} ^{\mathrm{1}} {k}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} {k}^{\mathrm{2}{p}} {dk} \\ $$$${u}\triangleq{k}^{\mathrm{2}} \Rightarrow{dk}=\frac{\mathrm{1}}{\mathrm{2}}{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} {du} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {k}^{\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{2}{p}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} {dk}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{\mathrm{1}}{\mathrm{6}}+{p}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} {u}^{−\frac{\mathrm{1}}{\mathrm{2}}} {du}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{p}−\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{3}}} {du} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{a}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{{b}−\mathrm{1}} {du}={B}\left({a},{b}\right)\:\mathrm{for}\:{a}>\mathrm{0},{b} \\ $$$${a}\triangleq{p}+\frac{\mathrm{2}}{\mathrm{3}},{b}\triangleq\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${B}\left({p}+\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{p}+\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} {du} \\ $$$$\Gamma\left({p}+\frac{\mathrm{3}}{\mathrm{4}}\right)=\left({p}+\frac{\mathrm{2}}{\mathrm{3}}\right)\Gamma\left({p}+\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$${B}\left({p}+\frac{\mathrm{2}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\Gamma\left({p}+\frac{\mathrm{2}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left({p}+\frac{\mathrm{3}}{\mathrm{4}}\right)}=\frac{\Gamma\left({p}+\frac{\mathrm{2}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\left({p}+\frac{\mathrm{2}}{\mathrm{3}}\right)\left({p}+\frac{\mathrm{2}}{\mathrm{3}}\right)}=\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{{p}+\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {k}^{\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{2}{p}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} {dk}=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{{p}+\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {k}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \boldsymbol{{K}}\left({k}\right)^{\mathrm{2}} {dk}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}} \frac{\mathrm{1}}{\mathrm{16}^{{p}} }\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{{p}+\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}} \frac{\mathrm{1}}{\mathrm{16}^{{p}} }\:\frac{\mathrm{1}}{{p}+\frac{\mathrm{2}}{\mathrm{3}}} \\ $$$$\frac{\mathrm{1}}{{p}+\frac{\mathrm{2}}{\mathrm{3}}}=\int_{\mathrm{0}} ^{\mathrm{1}} {v}^{{p}+\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} {dv}\:\forall{p}\in\mathbb{N} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}} \left(\frac{\mathrm{1}}{\mathrm{16}}\right)^{{p}} \frac{\mathrm{1}}{{p}+\frac{\mathrm{2}}{\mathrm{3}}}=\int_{\mathrm{0}} ^{\mathrm{1}} {v}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} \underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}} \left(\frac{{v}}{\mathrm{16}}\right)^{{p}} {dv} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}} {w}^{{p}} =\left(\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}^{\mathrm{2}} {w}^{{n}} \right)=\left(\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};\mathrm{16}{w}\right)\right)^{\mathrm{2}} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}\:} \left(\frac{\mathrm{1}}{\mathrm{16}}\right)^{{p}} +\frac{\mathrm{1}}{{p}+\frac{\mathrm{2}}{\mathrm{3}}}=\int_{\mathrm{0}} ^{\mathrm{1}} {v}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} \left(\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};\mathrm{16}\centerdot\frac{{v}}{\mathrm{16}}\right)\right)^{\mathrm{2}} {dv}=\int_{\mathrm{0}} ^{\mathrm{1}} {v}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} \left(\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1};{v}\right)\right)^{\mathrm{2}} {dv} \\ $$$$\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{1},{v}\right)=\frac{\mathrm{2}}{\pi}\boldsymbol{{K}}\left(\sqrt{{v}}\right)^{\mathrm{2}} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}} \left(\frac{\mathrm{1}}{\mathrm{16}}\right)^{{p}} \frac{\mathrm{1}}{{p}+\frac{\mathrm{2}}{\mathrm{3}}}=\int_{\mathrm{0}} ^{\mathrm{1}} {v}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} \frac{\mathrm{4}}{\pi^{\mathrm{2}} }\boldsymbol{{K}}\left(\sqrt{{v}}\right)^{\mathrm{2}} {dv} \\ $$$${t}\triangleq\sqrt{{v}}\Rightarrow{v}={t}^{\mathrm{2}} \Rightarrow{dv}=\mathrm{2}{tdt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {v}^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} \boldsymbol{{K}}\left(\sqrt{{v}}\right)^{\mathrm{2}} {dv}=\int_{\mathrm{0}} ^{\mathrm{1}} \left({t}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}−\mathrm{1}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} \mathrm{2}{tdt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{2}} −\boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {tdt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{c}_{{p}} \left(\frac{\mathrm{1}}{\mathrm{16}}\right)^{{p}} \frac{\mathrm{1}}{{p}+\frac{\mathrm{2}}{\mathrm{3}}}=\frac{\mathrm{4}}{\pi^{\mathrm{2}} }\centerdot\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt}=\frac{\mathrm{8}}{\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {k}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({k}\right)^{\mathrm{2}} {dk}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\centerdot\frac{\mathrm{8}}{\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt}=\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt} \\ $$$$\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} =\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\int_{\mathrm{0}\:} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt}=\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt} \\ $$$$\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt}=\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\centerdot\frac{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{{K}}\left({t}\right)^{\mathrm{2}} {dt} \\ $$$$\underline{\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\alpha−\mathrm{1}} }{{t}^{\pi} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{sin}\:\alpha}\:\mathrm{For}\:\mathrm{0}<\alpha<\pi} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{2}}{\pi}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{3}} \pi=\mathrm{2}\sqrt{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\frac{\pi}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{9}} =\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \\ $$$$\frac{\sqrt{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} }{\mathrm{64}\sqrt[{\mathrm{3}}]{\mathrm{2}}\pi^{\mathrm{3}} }=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{9}}\right)^{\mathrm{9}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{3}}\right)}{\mathrm{64}\sqrt[{\mathrm{3}}]{\mathrm{2}}\pi^{\mathrm{3}} }\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}\pi \\ $$$$\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{\mathrm{64}\sqrt[{\mathrm{3}}]{\mathrm{2}}\pi^{\mathrm{3}} \pi}\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)} \\ $$$$\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{128}\sqrt[{\mathrm{8}}]{\mathrm{2}}\pi^{\mathrm{4}} \Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} \sqrt{\mathrm{3}}}{\mathrm{256}\sqrt[{\mathrm{3}}]{\mathrm{2}}\pi^{\mathrm{4}} } \\ $$$$=\frac{\sqrt{\mathrm{3}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{9}} }{\mathrm{64}\sqrt[{\mathrm{3}}]{\mathrm{2}}\pi^{\mathrm{3}} } \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\sqrt[{\mathrm{4}}]{\mathrm{tan}\:{x}}}{\mathrm{sin}\:{x}}{dx}=\frac{\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{24}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right) \\ $$$$\left(\mathrm{1}\right):{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\:\mathrm{tan}^{\mathrm{1}/\mathrm{4}} {x}}{\mathrm{sin}\:{x}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\:\mathrm{tan}^{−\mathrm{3}/\mathrm{4}} {x}\:\mathrm{sec}\:{x}\:{dx} \\ $$$$\mathrm{tan}\:{x}={t}\Rightarrow{dx}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} },\mathrm{sec}\:{x}=\sqrt{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{arctan}{t}\centerdot{t}^{−\mathrm{3}/\mathrm{4}} }{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt} \\ $$$$\mathrm{arctan}\:{t}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}}{\mathrm{1}+{y}^{\mathrm{2}} {t}^{\mathrm{2}} }{dy} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+{y}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt}\:{dy} \\ $$$${t}=\frac{{u}}{{y}}\Rightarrow{dt}=\frac{{du}}{{y}},{t}^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} }{{y}^{\mathrm{2}} } \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{−\mathrm{5}/\mathrm{4}} \int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\sqrt{\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{y}^{\mathrm{2}} }}}{du}\:{dy} \\ $$$$\sqrt{\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{y}^{\mathrm{2}} }}=\sqrt{\frac{{y}^{\mathrm{2}} +{u}^{\mathrm{2}} }{{y}}} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {y}^{−\mathrm{1}/\mathrm{4}} \int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\sqrt{{y}^{\mathrm{2}} +{u}^{\mathrm{2}} }}{du}\:{dy} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\mathrm{1}/\mathrm{4}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\sqrt{{y}^{\mathrm{2}} +{u}^{\mathrm{2}} }}{dx}=\frac{\pi}{\mathrm{2}\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}}\left({y}^{−\mathrm{1}/\mathrm{4}} −{y}^{\mathrm{5}/\mathrm{4}} \right) \\ $$$${I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{y}^{−\mathrm{1}/\mathrm{4}} −{y}^{−\mathrm{5}/\mathrm{4}} }{\:\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}}{dy} \\ $$$${y}=\mathrm{sin}\theta\Rightarrow{dy}=\mathrm{cos}\:\theta\:{d}\theta \\ $$$${I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{−\mathrm{1}/\mathrm{4}} \theta−\mathrm{sin}^{\mathrm{5}/\mathrm{4}} \theta}{\:\sqrt{\mathrm{sin}\:\theta−\mathrm{1}}}\mathrm{cos}\:\theta\:{d}\theta \\ $$$$\sqrt{\mathrm{sin}^{\mathrm{2}} \theta−\mathrm{1}}={i}\:\mathrm{cos}\:\theta \\ $$$${I}=\frac{\pi}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}^{−\mathrm{1}/\mathrm{4}} \theta−\mathrm{sin}^{−\mathrm{5}/\mathrm{4}} \theta\right){d}\theta \\ $$$${I}=\frac{\pi}{\mathrm{2}{i}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}^{−\mathrm{1}/\mathrm{4}} −\mathrm{sin}^{−\mathrm{5}/\mathrm{4}} \theta\right){d}\theta \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{{a}} \theta\:{d}\theta=\frac{\sqrt{\pi}\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)} \\ $$$${I}=\frac{\pi}{\mathrm{2}{i}}\left[\frac{\sqrt{\pi}\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}−\frac{\sqrt{\pi}\Gamma\left(−\frac{\mathrm{1}}{\mathrm{8}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}\right] \\ $$$$\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{\pi}{\mathrm{sin}\frac{\pi}{\mathrm{8}}},\Gamma\left(−\frac{\mathrm{1}}{\mathrm{8}}\right)=−\frac{\mathrm{8}}{\mathrm{7}}\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right) \\ $$$${I}=\frac{\pi^{\mathrm{3}/\mathrm{2}} }{\mathrm{4}{i}}\left[\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)}{\pi}\mathrm{sin}\frac{\pi}{\mathrm{8}}+\frac{\mathrm{8}\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}{\mathrm{7}\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)}\right] \\ $$$$\mathrm{sin}\frac{\pi}{\mathrm{8}}=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$${I}=\frac{\pi^{\mathrm{3}/\mathrm{2}} }{\mathrm{4}{i}}\left[\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)}{\pi}\centerdot\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}+\frac{\mathrm{8}\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)}{\mathrm{7}}\right] \\ $$$$\Gamma\left(\frac{\mathrm{7}}{\mathrm{8}}\right)=\frac{\pi}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\mathrm{sin}\frac{\pi}{\mathrm{8}}} \\ $$$${I}=\frac{\pi^{\mathrm{3}/\mathrm{2}} }{\mathrm{4}{i}}\left[\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)}{\pi}\centerdot\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}+\frac{\mathrm{8}\pi}{\mathrm{7}\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\mathrm{sin}\frac{\pi}{\mathrm{8}}}\right] \\ $$$$\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{8}}},\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{8}}=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{24}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{8}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{8}}\right) \\ $$

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