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

## 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:

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

$\blacksquare$