Question Number 136060 by mnjuly1970 last updated on 18/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:\:{calculus}…. \\ $$$$\:\:\:\:{evaluate}:: \\ $$$$\:\:\:\:\boldsymbol{\phi}=\mathrm{Im}\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {li}_{\mathrm{2}} \left({sin}\left({x}\right)\right)+{li}_{\mathrm{2}} \left({csc}\left({x}\right)\right){dx}\right) \\ $$$$ \\ $$
Answered by mindispower last updated on 18/Mar/21
$${csc}\left({x}\right)=\frac{\mathrm{1}}{{sin}\left({x}\right)} \\ $$$${we}\:{have} \\ $$$${li}_{\mathrm{2}} \left({z}\right)+{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{z}}\right)=β\frac{\pi^{\mathrm{2}} }{\mathrm{6}}β\frac{{ln}^{\mathrm{2}} \left(β{z}\right)}{\mathrm{2}} \\ $$$$\Leftrightarrow\phi={Im}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(β\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}}β\frac{{ln}^{\mathrm{2}} \left(β{sin}\left({x}\right)\right)}{\mathrm{2}}\right){dx} \\ $$$$=β\frac{{Im}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({ln}\left({sin}\left({x}\right)\right)+{i}\pi\right)^{\mathrm{2}} {dx} \\ $$$$=β\frac{{Im}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{2}{i}\pi{ln}\left({sin}\left({x}\right)\right)\right){dx} \\ $$$$=β\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}=\frac{\pi^{\mathrm{2}} {log}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 18/Mar/21
$${thanks}\:{alot}\:{sir}\:{power}… \\ $$$$ \\ $$
Commented by mindispower last updated on 19/Mar/21
$${pleasur}\:{sir} \\ $$$$ \\ $$