Question Number 213776 by mnjuly1970 last updated on 16/Nov/24
$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{F}{ind}\:\:{the}\:\:{value}\:{of}\:\:{the}\:{following} \\ $$$$\:\:\:\:\:\:\:\:\:\:{expression}. \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\Omega=\:\:\:\frac{\:\mathrm{I}{m}\left(\:\mathrm{Li}_{\mathrm{2}} \:\left(\mathrm{2}\right)\right)}{\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\mathrm{ln}\left(\mathrm{sin}\left({x}\:\right)\right)\:{dx}}\:\:=\:? \\ $$
Answered by Frix last updated on 16/Nov/24
![∫_0 ^(π/2) ln sin x dx=−((πln 2)/2) Li_2 2 =−∫_0 ^2 ((ln (1−x))/x)dx =^([by parts]) =−[ln x ln (1−x)]_0 ^2 +∫_0 ^2 ((ln x)/(x−1))dx −[ln x ln (1−x)]_0 ^2 =−iπln 2 ∫_0 ^2 ((ln x)/(x−1))dx∈R ⇒ Ω=2](https://www.tinkutara.com/question/Q213783.png)
$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\mathrm{ln}\:\mathrm{sin}\:{x}\:{dx}=−\frac{\pi\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{Li}_{\mathrm{2}} \:\mathrm{2}\:=−\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:\frac{\mathrm{ln}\:\left(\mathrm{1}−{x}\right)}{{x}}{dx}\:\overset{\left[\mathrm{by}\:\mathrm{parts}\right]} {=} \\ $$$$=−\left[\mathrm{ln}\:{x}\:\mathrm{ln}\:\left(\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} +\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{ln}\:{x}}{{x}−\mathrm{1}}{dx} \\ $$$$−\left[\mathrm{ln}\:{x}\:\mathrm{ln}\:\left(\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} =−\mathrm{i}\pi\mathrm{ln}\:\mathrm{2} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\frac{\mathrm{ln}\:{x}}{{x}−\mathrm{1}}{dx}\in\mathbb{R} \\ $$$$\Rightarrow \\ $$$$\Omega=\mathrm{2} \\ $$
Commented by mnjuly1970 last updated on 16/Nov/24
$$\:{hello}\:{sir}\:{frix} \\ $$$${you}\:{are}\:{the}\:{best}\:. \\ $$$$\:\:\:{God}\:{bless}\:{you}… \\ $$
Commented by mnjuly1970 last updated on 16/Nov/24
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Commented by Frix last updated on 16/Nov/24
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{exactly}\:\mathrm{get}\:\mathrm{Li}_{\mathrm{2}} \:\mathrm{2}\:\mathrm{when}\:\mathrm{after}\:\mathrm{a}\:\mathrm{cup} \\ $$$$\mathrm{of}\:\mathrm{tea}\:\mathrm{I}\:\mathrm{suddenly}\:\mathrm{saw}\:\mathrm{that}\:\mathrm{the}\:\mathrm{remaining} \\ $$$$\mathrm{integral}\:\mathrm{was}\:>\mathrm{0}… \\ $$$$\Rightarrow\:\mathrm{my}\:\mathrm{advice}:\:\mathrm{Drink}\:\mathrm{more}\:\mathrm{tea}! \\ $$