Definite Integral from 0 to Pi of x by Logarithm of Sine x

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Theorem

$\ds \int_0^\pi x \map \ln {\sin x} \rd x = -\frac {\pi^2} 2 \ln 2$


Proof

\(\ds \int_0^\pi x \map \ln {\sin x} \rd x\) \(=\) \(\ds \int_0^\pi \paren {\pi - x} \map \ln {\map \sin {\pi - x} } \rd x\) Integral between Limits is Independent of Direction
\(\ds \) \(=\) \(\ds \pi \int_0^\pi \map \ln {\sin x} - \int_0^\pi x \map \ln {\sin x} \rd x\) Sine of Supplementary Angle, Linear Combination of Definite Integrals

So:

\(\ds 2 \int_0^\pi x \map \ln {\sin x} \rd x\) \(=\) \(\ds \pi \int_0^\pi \map \ln {\sin x} \rd x\)
\(\ds \) \(=\) \(\ds 2 \pi \int_0^{\pi/2} \map \ln {\sin x} \rd x\) Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Lemma
\(\ds \) \(=\) \(\ds -\pi^2 \ln 2\) Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$

giving:

$\ds \int_0^\pi x \map \ln {\sin x} \rd x = -\frac {\pi^2} 2 \ln 2$

$\blacksquare$


Sources

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