Find-the-value-of-the-following-expression-Im-Li-2-2-0-pi-2-ln-sin-x-dx-

Question Number 213776 by mnjuly1970 last updated on 16/Nov/24

F i n d t h e v a l u e o f t h e f o l l o w i n g

e x p r e s s i o n .

Ω = \frac{I m ({Li}_{2} (2))}{\int_{0}^{\frac{π}{2}} \ln (\sin (x)) d x} = ?

Answered by Frix last updated on 16/Nov/24

\underset{0}{\int^{\frac{π}{2}}} \ln \sin x d x = - \frac{π \ln 2}{2}

{Li}_{2} 2 = - \underset{0}{\int^{2}} \frac{\ln (1 - x)}{x} d x \overset{[by parts]}{=}

= - {[\ln x \ln (1 - x)]}_{0}^{2} + \underset{0}{\int^{2}} \frac{\ln x}{x - 1} d x

- {[\ln x \ln (1 - x)]}_{0}^{2} = - i π \ln 2

\underset{0}{\int^{2}} \frac{\ln x}{x - 1} d x \in R

\Rightarrow

Ω = 2

Commented by mnjuly1970 last updated on 16/Nov/24

h e l l o s i r f r i x

y o u a r e t h e b e s t .

G o d b l e s s y o u \dots

Commented by mnjuly1970 last updated on 16/Nov/24

Commented by Frix last updated on 16/Nov/24

I tried to exactly get {Li}_{2} 2 when after a cup

of tea I suddenly saw that the remaining

integral was > 0 \dots

\Rightarrow my advice : Drink more tea!

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