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

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Theorem

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


Proof

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

So:

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

giving:

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

$\blacksquare$


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