TITLE: Evaluate $\int_{0}^{\large\frac{\pi}{4}} \ln {(\sin x)}\cdot\ln {(\cos x)} \left(\frac{\ln{(\sin x)}}{\cot x}+\frac{\ln {(\cos x)}}{\tan x}\right)dx$
QUESTION [13 upvotes]: How do I find the value of this integral?
$$I=\int_{0}^{\Large\frac{\pi}{4}} \ln {(\sin x)}\cdot\ln {(\cos x)} \left(\dfrac{\ln{(\sin x)}}{\cot x}+\dfrac{\ln {(\cos x)}}{\tan x}\right)dx$$
I  tried substituting $t=\ln {(\sin x)}\cdot \ln {(\cos x)}$ and $t=\dfrac{\ln {(\sin x)}}{\ln {(\cos x)}}$, but it isn't working.

REPLY [4 votes]: Here I use the Beta function to evaluate this integral and the calculation is a little bit tedious. As Omran Kouba did, 
\begin{eqnarray}
I&=&\int_{0}^{\Large\frac{\pi}{2}} \ln^2 {(\sin x)}\cdot\ln {(\cos x)}\tan xdx.
\end{eqnarray}
Let
$$ J(a,b)=\int_0^{\frac{\pi}{2}}\sin^ax\cos^bx\tan xdx. $$
Then clearly
$$ \frac{\partial^3 J(0,0)}{\partial a^2\partial b}=I.$$
Now, by the change of variables $\sin x\to t$ and then $t^2\to u$, it is easy to see 
\begin{eqnarray}
J(a,b)=\frac{\Gamma(1+\frac{a}{2})\Gamma(\frac{b}{2})}{2\Gamma(1+\frac{a+b}{2})}.
\end{eqnarray}
Noting $\Gamma'(x)=\Gamma(x)\psi_0(x), \Gamma''(x)=\Gamma(x)\psi_0^2(x)+\Gamma(x)\psi_1(a)$, we have
\begin{eqnarray} \frac{\partial^2 J(a,b)}{\partial a^2}&=&\frac{\Gamma(1+\frac{a}{2})\Gamma(\frac{b}{2})}{8\Gamma(1+\frac{a+b}{2})}[\psi_0^2(1+\frac{a}{2})-2\psi_0(1+\frac{a}{2})\psi_0(1+\frac{a+b}{2})\\
&&+\psi_0^2(1+\frac{a+b}{2})+\psi_1(1+\frac{a}{2})-\psi_1(1+\frac{a+b}{2})]. 
\end{eqnarray}
Using
$$ \Gamma(1)=1, \psi_0(1)=\gamma, \psi_1(1)=\frac{\pi^2}{6}, $$
we obtain
$$ \frac{\partial^2 J(0,b)}{\partial a^2}=\frac{\Gamma(\frac{b}{2})}{48\Gamma(1+\frac{b}{2})}(6\gamma^2+\pi^2+12\gamma\psi_0(1+\frac{b}{2})+6\psi_0^2(1+\frac{b}{2})-6\psi_1(1+\frac{b}{2}))) $$
and hence
\begin{eqnarray}
\frac{\partial^3 J(0,b)}{\partial a^2\partial b}&=&\frac{\Gamma(\frac{b}{2})}{96\Gamma(1+\frac{b}{2})}[-6\psi_0^3(1+\frac{b}{2})+6\psi_0^2(1+\frac{b}{2})(-2\gamma+\psi_0(1+\frac{b}{2}))\\
&&+\psi_0(\frac{b}{2})(6\gamma^2+\pi^2-6\psi_1(1+\frac{b}{2}))+12\gamma\psi_1(1+\frac{b}{2})+\gamma\psi_0(1+\frac{b}{2})(-6\gamma^2-\pi^2\\
&&+12\gamma\psi_0(\frac{b}{2})+18\gamma\psi_1(1+\frac{b}{2}))-6\gamma\psi_2(1+\frac{b}{2})).
\end{eqnarray}
Noting, for $z\in(0,1)$,
$$ \Gamma(1)=1, \Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)} $$
we have
$$ \Gamma(\frac{b}{2})\approx\frac{\pi}{\sin(\frac{b\pi}{2})}, \psi_0(1+\frac{b}{2})=-\gamma+\frac{\pi^2b}{12}+O(b^2).$$
From this, we have
\begin{eqnarray}
\psi_0(\frac{b}{2})&=&-\frac{2}{b}-\gamma+\frac{\pi^2b}{12}+O(b^2)\\
\psi_0(1+\frac{b}{2})&=&-\gamma+\frac{\pi^2b}{12}+O(b^2)\\
\psi_1(1+\frac{b}{2})&=&\frac{\pi^2}{6}+\frac{b}{2}\psi_2(1)+\frac{\pi^4b^2}{120}+O(b^3)\\
\psi_2(1+\frac{b}{2})&=&\psi_2(1)+\frac{\pi^4b}{30}+O(b^2),
\end{eqnarray}
and hence
\begin{eqnarray}
\frac{\partial^3 J(0,b)}{\partial a^2\partial b}&\approx&\frac{b\pi^3}{11520\sin(\frac{b\pi}{2})}(-2\pi^2+b^2\pi^4+60b\psi_2(1)).
\end{eqnarray}
So
$$ \frac{\partial^3 J(0,0)}{\partial a^2\partial b}=\lim_{b\to0}\frac{\partial^3 J(0,b)}{\partial a^2\partial b}=\lim_{b\to0}\frac{b\pi^3}{11520\sin(\frac{b\pi}{2})}(-2\pi^2+b^2\pi^4+60b\psi_2(1))=-\frac{\pi^4}{2880}.$$