Q:

Evaluate $\int_{0}^{\large\frac{\pi}{4}} \ln {(\sin x)}\cdot\ln {(\cos x)} \left(\frac{\ln{(\sin x)}}{\cot x}+\frac{\ln {(\cos x)}}{\tan x}\right)dx$

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.

A:

We will prove that
$$I=-\frac{\pi^4}{2880}.$$
Indeed, let
$$
J=\int_0^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx
$$
It is easy to see that 
$$\eqalign{J&=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+
\int_{\pi/4}^{\pi/2}\log^2(\sin x)\log(\cos x)\tan x \,dx\cr
&=\int_0^{\pi/4}\log^2(\sin x)\log(\cos x)\tan x \,dx+
\int_{0}^{\pi/4}\log^2(\cos x)\log(\sin x)\cot x \,dx\cr
&=I
}$$
Now, to calculate $J$ we make the substitution $t\leftarrow\sin^2x$:
$$
J=\frac{1}{16}\int_0^1\frac{\log(1-u)}{1-u}\log^2(u)\,du
$$
But
$$\frac{\log(1-u)}{1-u}=-\left(\sum_{n=0}^\infty u^n\right)\left(\sum_{n=1}^\infty \frac{u^n}{n}\right)
=-\sum_{n=1}^\infty H_nu^n
$$
where $H_n=\sum_{k=1}^n1/k$. Hence
$$J=-\frac{1}{16}\sum_{n=1}^\infty H_n\int_0^1u^n\log^2(u)du
=-\frac{1}{8}\sum_{n=1}^\infty\frac{ H_n}{(n+1)^3} $$
Finally, since $H_{n}=H_{n+1}-\frac{1}{n+1}$, we get
$$J=\frac{1}{8}\zeta(4)-\frac{1}{8}\sum_{n=1}^\infty\frac{H_n}{n^3}\tag{1}$$
The sum $\sum_{n=1}^\infty\frac{H_n}{n^3}$ is known, it can be evaluated as follows, first we have
$$
H_n=\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+n}\right)=
\sum_{k=1}^\infty \frac{n}{k(k+n)} 
$$
Thus
$$
\sum_{n=1}^\infty\frac{H_n}{n^3}=\sum_{k,n\geq1}\frac{1}{n^2k(n+k)}
=\sum_{k,n\geq1}\frac{1}{k^2n(n+k)}
$$
Taking the half sum we find
$$
\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac{1}{2}\sum_{k,n\geq1}\frac{1}{kn(k+n)}\left(\frac{1}{k}+\frac{1}{n}\right)=
\frac{1}{2}\sum_{k,n\geq1}\frac{1}{k^2n^2}=\frac{1}{2}\zeta^2(2)
$$
Replacing in $(1)$ we obtain
$$J=\frac{1}{8}\zeta(4)-\frac{1}{16}\zeta^2(2)=-\frac{\pi^4}{2880}.$$
as announced.$\qquad\square$

A:

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
$$ \frac{\partial^3 J(0,0)}{\partial a^2\partial b}=\lim_{b\to0}\frac{\partial^3 J(0,b)}{\partial a^2\partial b}=\lim_{b\to0}\frac{b\pi^3}{11520\sin(\frac{b\pi}{2})}(-2\pi^2+b^2\pi^4+60b\psi_2(1))=-\frac{\pi^4}{2880}.$$

