TITLE: How to compute $\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x$
QUESTION [7 upvotes]: I am trying to compute this integral.
$$\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x$$
Any thoughts will help. Thanks.

REPLY [4 votes]: Note that
\begin{eqnarray}
&&\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x\\
&=&\int_0^{\frac{\pi}{2}}(\cos x)^{-1}\sin x\ln(\sin x)\ \text{d}x=\lim_{a\to0,b\to2}\frac{d}{db}\int_0^{\frac{\pi}{2}}(\cos x)^{a-1}(\sin x)^{b-1}\ \text{d}x\\
&=&\lim_{a\to0,b\to2}\frac{d}{db}\frac{1}{2}B\left(\frac{a}{2},\frac{b}{2}\right)\\
&=&\lim_{a\to0,b\to2}\frac{1}{4}B\left(\frac{a}{2},\frac{b}{2}\right)\left(\psi\left(\frac{b}{2}\right)-\psi\left(\frac{a+b}{2}\right)\right)\\
&=&\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\psi\left(\frac{b}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right)\\
&=&\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\psi(1)-\psi\left(\frac{a}{2}+1\right)\right)\\
&=&-\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\gamma+\psi\left(\frac{a}{2}+1\right)\right)\\
&=&-\lim_{a\to0}\frac{1}{4}\frac{\Gamma(\frac{a}{2})\Gamma(1)}{\Gamma(\frac{a+2}{2})}\left(\gamma+\psi\left(\frac{a}{2}+1\right)\right)\\
&=&-\lim_{a\to0}\frac{1}{4}\left(\frac{2}{a}+O(a^2)\right)\left(\frac{\pi^2}{12}a+O(a^2)\right)\\
&=&-\frac{\pi^2}{24}.
\end{eqnarray}
Here we use the following fact
$$ \Gamma(a)\approx\frac{1}{a}, \gamma+\psi(\frac{a}{2}+1)\approx\frac{\pi^2}{12}a.$$