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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))

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

Question Number 221523    Answers: 0   Comments: 1

Question Number 221521    Answers: 0   Comments: 1

Question Number 221513    Answers: 0   Comments: 0

Question Number 221541    Answers: 1   Comments: 0

∫_0 ^( π) tan(4!)^e^(3x^2 + 2x) .dx

$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{tan}\left(\mathrm{4}!\right)^{{e}^{\mathrm{3}{x}^{\mathrm{2}} +\:\mathrm{2}{x}} } .{dx} \\ $$

Question Number 221504    Answers: 1   Comments: 0

Find x if (x^4 /4^x )=(4^x /x^4 ) . x∈R

$${Find}\:{x}\:{if}\:\:\:\:\frac{{x}^{\mathrm{4}} }{\mathrm{4}^{{x}} }=\frac{\mathrm{4}^{{x}} }{{x}^{\mathrm{4}} }\:\:.\:\:\:{x}\in\mathbb{R} \\ $$

Question Number 221501    Answers: 3   Comments: 0

solve for x: (√(a−(√(a+x))))+(√(a+(√(a−x))))=2x it′s possible to solve for a but x seems impossible to me

$${solve}\:{for}\:\mathrm{x}: \\ $$$$\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}+\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}=\mathrm{2x} \\ $$$${it}'{s}\:{possible}\:{to}\:{solve}\:{for}\:\mathrm{a}\:{but}\:\mathrm{x}\:{seems} \\ $$$${impossible}\:{to}\:{me} \\ $$

Question Number 221500    Answers: 0   Comments: 0

Ip = ((p1q0)/(p0q0))×100

$$\boldsymbol{{Ip}}\:=\:\frac{\mathrm{p1q0}}{\mathrm{p0q0}}×\mathrm{100} \\ $$

Question Number 221498    Answers: 1   Comments: 0

Solve differantial Equation ((d )/dt)[((dy(t))/dt)]+ty(t)=0

$$\mathrm{Solve}\:\mathrm{differantial}\:\mathrm{Equation} \\ $$$$\frac{{d}\:\:}{{dt}}\left[\frac{{dy}\left({t}\right)}{{dt}}\right]+{ty}\left({t}\right)=\mathrm{0} \\ $$

Question Number 221495    Answers: 0   Comments: 1

Question Number 221484    Answers: 1   Comments: 0

I = ∫_( [0,1]^( 4) ) (1/( (√(1−x^2 y^2 −x^2 z^2 −x^2 w^2 −y^2 z^2 −y^2 w^2 −z^2 w^2 )))) dxdydzdw

$$ \\ $$$$\:\:\:{I}\:=\:\int_{\:\left[\mathrm{0},\mathrm{1}\right]^{\:\mathrm{4}} } \:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} {y}^{\mathrm{2}} −{x}^{\mathrm{2}} {z}^{\mathrm{2}} −{x}^{\mathrm{2}} {w}^{\mathrm{2}} −{y}^{\mathrm{2}} {z}^{\mathrm{2}} −{y}^{\mathrm{2}} {w}^{\mathrm{2}} −{z}^{\mathrm{2}} {w}^{\mathrm{2}} }}\:\mathrm{d}{x}\mathrm{d}{y}\mathrm{d}{z}\mathrm{d}{w}\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221483    Answers: 0   Comments: 1

I(α) = ∫∫∫_( [0,1]^3 ) ((ln(1 + α(xy + yz + zx)))/(1 − xyz )) dxdydz for α ∈ (0,1)

$$ \\ $$$$\:\:\:\:{I}\left(\alpha\right)\:=\:\int\int\int_{\:\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } \:\frac{\mathrm{ln}\left(\mathrm{1}\:+\:\alpha\left({xy}\:+\:{yz}\:+\:{zx}\right)\right)}{\mathrm{1}\:−\:{xyz}\:}\:{dxdydz}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\alpha\:\in\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$$ \\ $$

Question Number 221481    Answers: 0   Comments: 0

prove:Σ_(k=0) ^n ((k!(n−k)!)/(n!))=−((i2^(n−1) Γ(n+2)(π−iB_2 (n+2,0)))/(n!))

$$\mathrm{prove}:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{k}!\left({n}−{k}\right)!}{{n}!}=−\frac{{i}\mathrm{2}^{{n}−\mathrm{1}} \Gamma\left({n}+\mathrm{2}\right)\left(\pi−{iB}_{\mathrm{2}} \left({n}+\mathrm{2},\mathrm{0}\right)\right)}{{n}!} \\ $$

Question Number 221469    Answers: 1   Comments: 0

Prove; ∫_0 ^( 1) ((((√(x )) ln (√x))^2 )/((1−x^2 )^2 + 4x^2 + 2x^4 + 3x^6 + 2x^8 + x^(10) )) dx = ((21)/(1024)) ζ(3) − (π^3 /(1024)) + (π^3 /(324(√3)))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}; \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\left(\sqrt{{x}\:}\:\mathrm{ln}\:\sqrt{{x}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}^{\mathrm{4}} \:+\:\mathrm{3}{x}^{\mathrm{6}} \:+\:\mathrm{2}{x}^{\mathrm{8}} \:+\:{x}^{\mathrm{10}} }\:\:\mathrm{d}{x} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{21}}{\mathrm{1024}}\:\zeta\left(\mathrm{3}\right)\:−\:\frac{\pi^{\mathrm{3}} }{\mathrm{1024}}\:+\:\frac{\pi^{\mathrm{3}} }{\mathrm{324}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Question Number 221454    Answers: 2   Comments: 0

Prove ∫^( +∞) _( 1) (x/((−1 + 3x^2 − 3x^4 + 2x^6 )(ln(x−1) − 2 ln x + ln(x+1)))) dx = − ((ln 3)/4)

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove} \\ $$$$\:\:\underset{\:\mathrm{1}} {\int}^{\:+\infty} \:\frac{{x}}{\left(−\mathrm{1}\:+\:\mathrm{3}{x}^{\mathrm{2}} \:−\:\mathrm{3}{x}^{\mathrm{4}} \:+\:\mathrm{2}{x}^{\mathrm{6}} \right)\left(\mathrm{ln}\left({x}−\mathrm{1}\right)\:−\:\mathrm{2}\:\mathrm{ln}\:{x}\:+\:\mathrm{ln}\left({x}+\mathrm{1}\right)\right)}\:\:\mathrm{d}{x}\:=\:−\:\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221453    Answers: 2   Comments: 0

Question Number 221447    Answers: 1   Comments: 0

if Π_(i=1) ^n (x + r_i ) ≡ Σ_(j=0) ^n a_j x^(n−i) show that ; Σ_(i=1) ^n tan^(−1) r_i = tan^(−1) ((a_1 + a_3 + a_5 − ∙∙∙)/(a_0 − a_2 + a_4 −∙∙∙)) and Σ_(i=1) ^n tanh^(−1) r_i = tanh^(−1) ((a_1 + a_3 + a_5 + ∙∙∙)/(a_0 + a_2 + a_4 + ∙∙∙))

$$ \\ $$$$\:\:\:\:\mathrm{if}\:\:\:\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\:\left({x}\:+\:{r}_{{i}} \right)\:\equiv\:\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}\:{a}_{{j}} {x}^{{n}−{i}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{show}\:\mathrm{that}\:; \\ $$$$\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{tan}^{−\mathrm{1}} \:{r}_{{i}} \:=\:\mathrm{tan}^{−\mathrm{1}} \:\frac{{a}_{\mathrm{1}} +\:{a}_{\mathrm{3}} \:+\:{a}_{\mathrm{5}} \:−\:\centerdot\centerdot\centerdot}{{a}_{\mathrm{0}} \:−\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{4}} \:−\centerdot\centerdot\centerdot}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{and} \\ $$$$\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{tanh}^{−\mathrm{1}} \:{r}_{{i}} \:=\:\mathrm{tanh}^{−\mathrm{1}} \:\frac{{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{3}} \:+\:{a}_{\mathrm{5}} \:+\:\centerdot\centerdot\centerdot}{{a}_{\mathrm{0}} \:+\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{4}} \:+\:\centerdot\centerdot\centerdot}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221444    Answers: 0   Comments: 3

Question Number 221438    Answers: 0   Comments: 0

∫_( 1) ^∞ ((ln(ln x))/(1−2x cos θ + x^2 )) dx ; for all θ ∈ (−π , π)

$$ \\ $$$$\:\:\:\int_{\:\mathrm{1}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)}{\mathrm{1}−\mathrm{2}{x}\:\mathrm{cos}\:\theta\:+\:{x}^{\mathrm{2}} }\:{dx}\:;\:\mathrm{for}\:\mathrm{all}\:\theta\:\in\:\left(−\pi\:,\:\pi\right)\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221430    Answers: 1   Comments: 0

Question Number 221417    Answers: 0   Comments: 0

Let S be delimited by the equations x=0; y=0 ; z=0 and x+y+z=0 Find the flux of vector field V(x,y,z)=(x,y,x^2 +y^2 ) through S

$${Let}\:\:{S}\:{be}\:{delimited}\:{by}\:{the}\:{equations}\: \\ $$$${x}=\mathrm{0};\:{y}=\mathrm{0}\:;\:{z}=\mathrm{0}\:{and}\:{x}+{y}+{z}=\mathrm{0} \\ $$$${Find}\:{the}\:{flux}\:{of}\:{vector}\:{field}\: \\ $$$${V}\left({x},{y},{z}\right)=\left({x},{y},{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:{through}\:{S} \\ $$

Question Number 221416    Answers: 0   Comments: 0

ex3. prove f^((n)) (α)=((n!)/(2πi)) ∮_( ∂S) ((f(z))/((z−α)^(n+1) )) dz ex4. Let z_0 be any point interior to a positively oriented simple closed contour C show that a. ∮_C (dz/(z−z_0 ))=2πi b. ∮_( C) (dz/((z−z_0 )^(n+1) ))=0 , n∈R^+ ex 5. Let C be any simple closed contour, described in the positive sense in the z plane and write g(z)= ∮_( C) ((s^3 +2s)/((s−z)^3 )) ds show that g(z)=6πi when z is inside C and that g(z)=0 when z is outside

$$\mathrm{ex3}. \\ $$$$\mathrm{prove} \\ $$$${f}^{\left({n}\right)} \left(\alpha\right)=\frac{{n}!}{\mathrm{2}\pi\boldsymbol{{i}}}\:\oint_{\:\partial{S}} \:\frac{{f}\left({z}\right)}{\left({z}−\alpha\right)^{{n}+\mathrm{1}} }\:\mathrm{d}{z} \\ $$$$\mathrm{ex4}. \\ $$$$\mathrm{Let}\:{z}_{\mathrm{0}} \:\mathrm{be}\:\mathrm{any}\:\mathrm{point}\:\mathrm{interior}\:\mathrm{to}\:\mathrm{a}\:\mathrm{positively} \\ $$$$\mathrm{oriented}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}\:\mathcal{C} \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$${a}.\:\oint_{{C}} \:\frac{\mathrm{d}{z}}{{z}−{z}_{\mathrm{0}} }=\mathrm{2}\pi\boldsymbol{{i}} \\ $$$$\mathrm{b}.\:\oint_{\:{C}} \frac{\mathrm{d}{z}}{\left({z}−{z}_{\mathrm{0}} \right)^{{n}+\mathrm{1}} }=\mathrm{0}\:,\:{n}\in\mathbb{R}^{+} \\ $$$$\mathrm{ex}\:\mathrm{5}. \\ $$$$\mathrm{Let}\:\mathcal{C}\:\mathrm{be}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}, \\ $$$$\mathrm{described}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{sense}\:\mathrm{in}\:\mathrm{the}\:{z}\:\mathrm{plane} \\ $$$$\mathrm{and}\:\mathrm{write}\: \\ $$$$\mathrm{g}\left({z}\right)=\:\oint_{\:\mathcal{C}} \:\frac{{s}^{\mathrm{3}} +\mathrm{2}{s}}{\left({s}−{z}\right)^{\mathrm{3}} }\:\mathrm{d}{s} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{6}\pi\boldsymbol{{i}}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{inside}\:\mathcal{C}\:\mathrm{and} \\ $$$$\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{0}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{outside} \\ $$

Question Number 221415    Answers: 1   Comments: 0

∫ dz [−(1/π)((2/z))^ν ∙Σ_(k=0) ^(ν−1) ((Γ(ν−k))/(k!))((z/2))^(2k) +(2/π)ln((1/2)z)J_ν (z)−(1/π)((z/2))^ν Σ_(k=0) ^∞ (((−1)^k (ψ^((0)) (k+ν+1)+ψ^((0)) (k+1)))/(k!(k+ν)!))((z/2))^(2k) ]

$$\int\:\:\mathrm{d}{z}\:\left[−\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \centerdot\underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\:\frac{\Gamma\left(\nu−{k}\right)}{{k}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\frac{\mathrm{2}}{\pi}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{z}\right){J}_{\nu} \left({z}\right)−\frac{\mathrm{1}}{\pi}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} \left(\psi^{\left(\mathrm{0}\right)} \left({k}+\nu+\mathrm{1}\right)+\psi^{\left(\mathrm{0}\right)} \left({k}+\mathrm{1}\right)\right)}{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \right] \\ $$

Question Number 221413    Answers: 7   Comments: 0

Question Number 221411    Answers: 1   Comments: 0

Question Number 221407    Answers: 0   Comments: 0

A and B are two angles such that 0^0 <B<A<90^0 then prove geometrycaly that cos (A+B)=cos Acos B−sin Asin B

$${A}\:{and}\:{B}\:{are}\:{two}\:{angles}\:{such}\:{that}\:\mathrm{0}^{\mathrm{0}} <{B}<{A}<\mathrm{90}^{\mathrm{0}} \:{then}\:{prove}\:{geometrycaly}\:{that}\: \\ $$$$\mathrm{cos}\:\left({A}+{B}\right)=\mathrm{cos}\:{A}\mathrm{cos}\:{B}−\mathrm{sin}\:{A}\mathrm{sin}\:{B} \\ $$

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