Definite Integral from 0 to Half Pi of Logarithm of Sine x/Proof 2

Proof
By Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$: Lemma, we have:


 * $\displaystyle \int_0^\pi \map \ln {\sin x} \rd x = 2 \int_0^{\pi/2} \map \ln {\sin x} \rd x$

By Product of Sines of Fractions of Pi, we have:


 * $\displaystyle \prod_{k \mathop = 1}^{n - 1} \map \sin {\frac {k \pi} n} = \frac n {2^{n - 1} }$

Therefore, we have:

We have:

We can show the first term to vanish:

So:

giving:


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