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Question Number 136060 by mnjuly1970 last updated on 18/Mar/21

...advanced calculus.... evaluate:: 𝛗=Im(∫_0 ^( (Ο€/2)) li_2 (sin(x))+li_2 (csc(x))dx)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{advanced}\:\:\:{calculus}.... \\ $$$$\:\:\:\:{evaluate}:: \\ $$$$\:\:\:\:\boldsymbol{\phi}=\mathrm{Im}\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {li}_{\mathrm{2}} \left({sin}\left({x}\right)\right)+{li}_{\mathrm{2}} \left({csc}\left({x}\right)\right){dx}\right) \\ $$$$ \\ $$

Answered by mindispower last updated on 18/Mar/21

csc(x)=(1/(sin(x))) we have li_2 (z)+li_2 ((1/z))=βˆ’(Ο€^2 /6)βˆ’((ln^2 (βˆ’z))/2) ⇔φ=Im∫_0 ^(Ο€/2) (βˆ’(𝛑^2 /6)βˆ’((ln^2 (βˆ’sin(x)))/2))dx =βˆ’((Im)/2)∫_0 ^(Ο€/2) (ln(sin(x))+iΟ€)^2 dx =βˆ’((Im)/2)∫_0 ^(Ο€/2) (2iΟ€ln(sin(x)))dx =βˆ’Ο€βˆ«_0 ^(Ο€/2) ln(sin(x))dx=((Ο€^2 log(2))/2)

$${csc}\left({x}\right)=\frac{\mathrm{1}}{{sin}\left({x}\right)} \\ $$$${we}\:{have} \\ $$$${li}_{\mathrm{2}} \left({z}\right)+{li}_{\mathrm{2}} \left(\frac{\mathrm{1}}{{z}}\right)=βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{6}}βˆ’\frac{{ln}^{\mathrm{2}} \left(βˆ’{z}\right)}{\mathrm{2}} \\ $$$$\Leftrightarrow\phi={Im}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(βˆ’\frac{\boldsymbol{\pi}^{\mathrm{2}} }{\mathrm{6}}βˆ’\frac{{ln}^{\mathrm{2}} \left(βˆ’{sin}\left({x}\right)\right)}{\mathrm{2}}\right){dx} \\ $$$$=βˆ’\frac{{Im}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({ln}\left({sin}\left({x}\right)\right)+{i}\pi\right)^{\mathrm{2}} {dx} \\ $$$$=βˆ’\frac{{Im}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{2}{i}\pi{ln}\left({sin}\left({x}\right)\right)\right){dx} \\ $$$$=βˆ’\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}=\frac{\pi^{\mathrm{2}} {log}\left(\mathrm{2}\right)}{\mathrm{2}} \\ $$

Commented by mnjuly1970 last updated on 18/Mar/21

thanks alot sir power...

$${thanks}\:{alot}\:{sir}\:{power}... \\ $$$$ \\ $$

Commented by mindispower last updated on 19/Mar/21

pleasur sir

$${pleasur}\:{sir} \\ $$$$ \\ $$

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