Question and Answers Forum

[ 1 .] ∫_0 ^( 1) ((ln(x) ln(1−x^2 )ln(1+x^2 ))/(1−x^2 )) dx [ 2 .] ∫_0 ^1 ((ln(x) ln(1−x) ln(1+x) ln(1+x^2 ))/(1+x)) dx [ 3 .] ∫_0 ^1 ((ln(x) ln(1−x^2 ) ln(1+x^2 ))/x) dx [ 4 .] ∫_0 ^( 1) ((ln(x) ln(1−x) ln(1+x) ln(1−x^2 ))/x) dx

Question Number 222409 Answers: 0 Comments: 0

Solve ; ∫_0 ^(π/2) ((ln^n sin θ)/(sin^p θ cos^q θ)) dθ , for n,p,q ∈ R_(≥ 0)

$$ \\ $$$$\:\:\:\mathrm{Solve}\:;\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{ln}^{{n}} \:\mathrm{sin}\:\theta}{\mathrm{sin}^{{p}} \:\theta\:\mathrm{cos}^{{q}} \:\theta}\:\mathrm{d}\theta\:,\:\mathrm{for}\:{n},{p},{q}\:\in\:\mathbb{R}_{\geqslant\:\mathrm{0}} \:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 222408 Answers: 1 Comments: 0

∫_1 ^∞ ((1/x))^(x/( (√(x−1)))) dx = ??

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{1}} ^{\infty} \:\left(\frac{\mathrm{1}}{{x}}\right)^{\frac{{x}}{\:\sqrt{{x}−\mathrm{1}}}} \:{dx}\:=\:\:\:?? \\ $$$$ \\ $$

Question Number 222336 Answers: 0 Comments: 0

∫∫∫ ((−y ± (√(y^2 + 4xy)))/(2x)) dxdydz

$$ \\ $$$$\:\:\:\:\:\:\:\int\int\int\:\:\frac{−{y}\:\pm\:\sqrt{{y}^{\mathrm{2}} \:+\:\mathrm{4}{xy}}}{\mathrm{2}{x}}\:{dxdydz} \\ $$$$ \\ $$

Question Number 222331 Answers: 0 Comments: 0

Find closed form; ∫_( 0) ^( 1) ((Li_2 (z^2 )Li_2 (−z^2 ))/(1 + z^2 )) dz = ?

$$ \\ $$$$\:\:\:\:\:\:\:\mathrm{Find}\:\mathrm{closed}\:\mathrm{form}; \\ $$$$\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{Li}_{\mathrm{2}} \left({z}^{\mathrm{2}} \right)\mathrm{Li}_{\mathrm{2}} \left(−{z}^{\mathrm{2}} \right)}{\mathrm{1}\:+\:{z}^{\mathrm{2}} }\:\mathrm{d}{z}\:=\:? \\ $$

Question Number 222295 Answers: 1 Comments: 0

Question Number 222292 Answers: 1 Comments: 0

∫_(−∞) ^∞ sech(z) sech(z−a) dz

$$ \\ $$$$\:\:\:\int_{−\infty} ^{\infty} \mathrm{sech}\left({z}\right)\:\mathrm{sech}\left({z}−{a}\right)\:{dz} \\ $$$$ \\ $$

Question Number 222275 Answers: 1 Comments: 0

Prove ; ∫_(−π) ^( π) ((z sin(z) )/((1 + z + (√(1 + z^2 )))(√(3 + sin^2 (z))))) dz = ζ(2)

$$ \\ $$$$\:\:\mathrm{Prove}\:;\:\int_{−\pi} ^{\:\pi} \:\frac{{z}\:\mathrm{sin}\left({z}\right)\:}{\left(\mathrm{1}\:+\:{z}\:+\:\sqrt{\mathrm{1}\:+\:{z}^{\mathrm{2}} }\right)\sqrt{\mathrm{3}\:+\:\mathrm{sin}^{\mathrm{2}} \left({z}\right)}}\:{dz}\:=\:\zeta\left(\mathrm{2}\right)\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 222245 Answers: 2 Comments: 0

∫_0 ^( (π/2)) ((cos^(−1) (((√(1 − sin^2 (x) cos^2 (x)))/(1 + sin^2 (x))))∙ln(((1 + sin(x))/(1 + cos(x)))))/( (√(1 + cos^2 (x) − sin^2 (x))))) dx

$$ \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{cos}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)\:\mathrm{cos}^{\mathrm{2}} \left({x}\right)}}{\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)}\right)\centerdot\mathrm{ln}\left(\frac{\mathrm{1}\:+\:\mathrm{sin}\left({x}\right)}{\mathrm{1}\:+\:\mathrm{cos}\left({x}\right)}\right)}{\:\sqrt{\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \left({x}\right)\:−\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)}}\:\:\mathrm{d}{x}\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 222224 Answers: 0 Comments: 1

Prove:∫_0 ^1 ((ln(1−x^2 ))/x)cos(ln x)dx=1−(π/2)cosh(π/2)

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\mathrm{cos}\left(\mathrm{ln}\:{x}\right){dx}=\mathrm{1}−\frac{\pi}{\mathrm{2}}\mathrm{cosh}\frac{\pi}{\mathrm{2}} \\ $$

Question Number 222218 Answers: 1 Comments: 0

∫_0 ^( π) tan^(−1) (((ln sin(x))/x)) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\mathrm{ln}\:\mathrm{sin}\left({x}\right)}{{x}}\right)\:{dx} \\ $$$$ \\ $$

Question Number 222217 Answers: 0 Comments: 0

∫_0 ^∞ ∫_0 ^∞ ((ln x ln y)/( (√(xy))))cos(x+y)=π^2 (γ+2 ln 2) Sol:∫_0 ^∞ ∫_0 ^∞ ((ln x ln y)/( (√(xy))))cos(x+y)dxdy=Re((∫_0 ^∞ x^(−(1/2)) e^(ix) ln xdx)(∫_0 ^∞ y^(−(1/2)) e^(iy) ln ydy)) ∫_0 ^∞ x^a e^(ix) dx=e^(iπ(a+1)) Γ(a+1),−1<Re a<0 ∫_0 ^∞ x^a e^(ix) dx=∫_0 ^∞ x^a e^(ix) ln xdx=(∂/∂u)[e^(iπ(a+1)/2) Γ(a+1)] =e^(iπ(a+1)) Γ(a+1)(((iπ)/2)+ψ(a+1)) a=−(1/2): ∫_0 ^∞ x^(−(1/2)) e^(ix) ln xdx=e^(iπ/1) (√π)(((iπ)/2)+ψ((1/2))) c=e^(iπ/4) (√π)(((iπ)/2)−γ−2 ln 2) e^(iπ/4) =((√2)/2)(1+i) c=(√π)∙((√2)/2)(1+i)(−γ−2 ln 2+i(π/2))=((√(2π))/2)[(γ−ln 2−(π/2))+i(−γ−2 ln 2+(π/2))] p=−γ−2 ln 2−(π/2),q=−γ−2 ln 2+(π/2) c^2 =(((√(2π))/2))^2 (p+ip)^2 =((2π)/4)(p^2 −q^2 +2ipq)=(π/2)(p^2 −q^2 +2ipq) p+q+2=(−γ−2 ln 2) p−q=π p^2 −q^2 =(p−q)(p+q)=(−π)∙2(−γ−2 ln 2)=2π(γ+2 ln 2) [2π(γ+2 ln 2)+2ipq]=π^2 (γ+2 ln 2)+iπpq Re(c^2 )=π^2 (γ+2 ln 2)

Question Number 222175 Answers: 1 Comments: 0

solve (e^(2y) −y)cosx(dy/dx)=e^y sin2x klipto−quanta

$$\boldsymbol{\mathrm{solve}} \\ $$$$\left(\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{y}}} −\boldsymbol{\mathrm{y}}\right)\boldsymbol{\mathrm{cosx}}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}} \\ $$

Question Number 222127 Answers: 0 Comments: 0

∫_0 ^( ∞) ((ln(tan(tan^(−1) (e^((1/π) tan^(−1) u) ))) )/(u^2 + 2πu + 2π^2 )) du

$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{tan}^{−\mathrm{1}} \left({e}^{\frac{\mathrm{1}}{\pi}\:\mathrm{tan}^{−\mathrm{1}} \:{u}} \right)\right)\right)\:}{{u}^{\mathrm{2}} \:+\:\mathrm{2}\pi{u}\:+\:\mathrm{2}\pi^{\mathrm{2}} }\:{du} \\ $$$$ \\ $$

Question Number 222100 Answers: 0 Comments: 0