Question Number 221843 by wewji12 last updated on 11/Jun/25
![∫_0 ^( (π/2)) x^2 csc^2 (x)dx ∫ −((4z^2 )/((e^(iz) −e^(−iz) )^2 )) dz=∫ ((z^2 e^(2iz) )/((e^(2iz) −1)^2 )) dz u=^(Substitute) e^(2iz) −1 →du=2ie^(2iz) dz z^2 =−((ln^2 (u+1))/4) −(1/8)i ∫ ((ln^2 (u+1))/u^2 ) du by parts ∫ f(u)g′(u)du=f(u)g(u)−∫ f′(u)g(u)du f(u)=ln^2 (u+1) g′(u)=(1/u) f′(u)=((2ln(u+1))/(u+1)) g(u)=−(1/u) −((ln^2 (u+1))/u)−∫ −((2ln(u+1))/(u(u+1))) du −2∫ ((ln(u+1))/(u(u+1))) du...? ∫ ( ((ln(u+1))/u)−((ln(u+1))/(u+1)))du ∫ ((ln(u+1))/u) du=−∫ −((ln(1−v))/v) dv ∴−Li_2 (−u) ∫ ((ln(u+1))/(u+1)) du=∫ v dv (∵ ln(u+1)=v) (1/2)(ln(u+1))^2 ∫ ((ln(u+1))/u) du−∫ ((ln(u+1))/(u+1)) du= −((ln^2 (u+1))/2)−Li_2 (−u) −2∫ ((ln(u+1))/(u(u+1))) du=ln^2 (u+1)+2Li_2 (−u) −((ln^2 (u+1))/u)−∫ −((2ln(u+1))/(u+1)) du= −((ln^2 (u+1))/u)−ln^2 (u+1)−2Li_2 (−u) plug to solve integrals (1/8)i ∫ ((ln^2 (u+1))/u^2 ) du −((iln^2 (u+1))/(8u))−((iln^2 (u+1))/8)−((iLi_2 (−u))/4) u=e^(2ix) −1 ∴ iLi_2 (1−e^(2ix) )−((2ix^2 )/(e^(2ix) −1))−2ix^2 +Const ∫_0 ^( (π/2)) - =[iLi_2 (1−e^(2ix) )−((2ix^2 )/(e^(2ix) −1))−2ix^2 ]_(x=0) ^(x=(π/2))](Q221843.png)
$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\:{x}^{\mathrm{2}} \mathrm{csc}^{\mathrm{2}} \left({x}\right)\mathrm{d}{x} \\ $$$$\int\:\:−\frac{\mathrm{4}{z}^{\mathrm{2}} }{\left({e}^{\boldsymbol{{i}}{z}} −{e}^{−\boldsymbol{{i}}{z}} \right)^{\mathrm{2}} }\:\mathrm{d}{z}=\int\:\:\frac{{z}^{\mathrm{2}} {e}^{\mathrm{2}\boldsymbol{{i}}{z}} }{\left({e}^{\mathrm{2}\boldsymbol{{i}}{z}} −\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{d}{z} \\ $$$${u}\overset{\mathrm{Substitute}} {=}{e}^{\mathrm{2}\boldsymbol{{i}}{z}} −\mathrm{1}\:\:\rightarrow\mathrm{d}{u}=\mathrm{2}\boldsymbol{{i}}{e}^{\mathrm{2}\boldsymbol{{i}}{z}} \:\mathrm{d}{z} \\ $$$${z}^{\mathrm{2}} =−\frac{\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{\mathrm{4}}\: \\ $$$$−\frac{\mathrm{1}}{\mathrm{8}}\boldsymbol{{i}}\:\int\:\:\frac{\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{{u}^{\mathrm{2}} }\:\mathrm{d}{u} \\ $$$$\mathrm{by}\:\mathrm{parts}\:\int\:{f}\left({u}\right)\mathrm{g}'\left({u}\right)\mathrm{d}{u}={f}\left({u}\right)\mathrm{g}\left({u}\right)−\int\:{f}'\left({u}\right)\mathrm{g}\left({u}\right)\mathrm{d}{u} \\ $$$${f}\left({u}\right)=\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)\:\mathrm{g}'\left({u}\right)=\frac{\mathrm{1}}{{u}} \\ $$$${f}'\left({u}\right)=\frac{\mathrm{2ln}\left({u}+\mathrm{1}\right)}{{u}+\mathrm{1}}\:\mathrm{g}\left({u}\right)=−\frac{\mathrm{1}}{{u}} \\ $$$$−\frac{\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{{u}}−\int\:−\frac{\mathrm{2ln}\left({u}+\mathrm{1}\right)}{{u}\left({u}+\mathrm{1}\right)}\:\mathrm{d}{u} \\ $$$$−\mathrm{2}\int\:\:\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}\left({u}+\mathrm{1}\right)}\:\mathrm{d}{u}...? \\ $$$$\int\:\left(\:\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}}−\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}+\mathrm{1}}\right)\mathrm{d}{u} \\ $$$$\int\:\:\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}}\:\mathrm{d}{u}=−\int\:−\frac{\mathrm{ln}\left(\mathrm{1}−{v}\right)}{{v}}\:\mathrm{d}{v} \\ $$$$\therefore−\mathrm{Li}_{\mathrm{2}} \left(−{u}\right) \\ $$$$\int\:\:\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}+\mathrm{1}}\:\mathrm{d}{u}=\int\:{v}\:\mathrm{d}{v}\:\left(\because\:\mathrm{ln}\left({u}+\mathrm{1}\right)={v}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\left({u}+\mathrm{1}\right)\right)^{\mathrm{2}} \\ $$$$\int\:\:\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}}\:\mathrm{d}{u}−\int\:\:\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}+\mathrm{1}}\:\mathrm{d}{u}= \\ $$$$−\frac{\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{\mathrm{2}}−\mathrm{Li}_{\mathrm{2}} \left(−{u}\right) \\ $$$$−\mathrm{2}\int\:\:\frac{\mathrm{ln}\left({u}+\mathrm{1}\right)}{{u}\left({u}+\mathrm{1}\right)}\:\mathrm{d}{u}=\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)+\mathrm{2Li}_{\mathrm{2}} \left(−{u}\right) \\ $$$$−\frac{\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{{u}}−\int\:\:−\frac{\mathrm{2ln}\left({u}+\mathrm{1}\right)}{{u}+\mathrm{1}}\:\mathrm{d}{u}= \\ $$$$−\frac{\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{{u}}−\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)−\mathrm{2Li}_{\mathrm{2}} \left(−{u}\right) \\ $$$$\mathrm{plug}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{integrals} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\boldsymbol{{i}}\:\int\:\:\frac{\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{{u}^{\mathrm{2}} }\:\mathrm{d}{u} \\ $$$$−\frac{\boldsymbol{{i}}\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{\mathrm{8}{u}}−\frac{\boldsymbol{{i}}\mathrm{ln}^{\mathrm{2}} \left({u}+\mathrm{1}\right)}{\mathrm{8}}−\frac{\boldsymbol{{i}}\mathrm{Li}_{\mathrm{2}} \left(−{u}\right)}{\mathrm{4}} \\ $$$${u}={e}^{\mathrm{2}\boldsymbol{{i}}{x}} −\mathrm{1} \\ $$$$\therefore\:\boldsymbol{{i}}\mathrm{Li}_{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}\boldsymbol{{i}}{x}} \right)−\frac{\mathrm{2}\boldsymbol{{i}}{x}^{\mathrm{2}} }{{e}^{\mathrm{2}\boldsymbol{{i}}{x}} −\mathrm{1}}−\mathrm{2}\boldsymbol{{i}}{x}^{\mathrm{2}} +\mathrm{Const} \\ $$$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:-\:=\left[\boldsymbol{{i}}\mathrm{Li}_{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}\boldsymbol{{i}}{x}} \right)−\frac{\mathrm{2}\boldsymbol{{i}}{x}^{\mathrm{2}} }{{e}^{\mathrm{2}\boldsymbol{{i}}{x}} −\mathrm{1}}−\mathrm{2}\boldsymbol{{i}}{x}^{\mathrm{2}} \right]_{{x}=\mathrm{0}} ^{{x}=\frac{\pi}{\mathrm{2}}} \\ $$
Commented by wewji12 last updated on 11/Jun/25
$$\mathrm{Q221838} \\ $$
Commented by Frix last updated on 11/Jun/25
$$\mathrm{I}\:\mathrm{found}\:\mathrm{another}\:\mathrm{path}.\:\mathrm{See}\:\mathrm{original}\:\mathrm{question}. \\ $$