<s>[INST] Twist on log of sine and cosine integral $\int_{0}^{\frac{\pi}{2}}x\ln(\sin x)\ln(\cos x)dx$ I ran across this integral and have not been able to evaluate it.
$$\displaystyle \int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))\ln(\cos(x))dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}-\frac{{\pi}^{4}}{192}$$
I had some ideas.  Perhaps some how arrive at $\displaystyle\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{4}}=\frac{{\pi}^{4}}{192}$.
and $\displaystyle \ln(2)\int_{0}^{\frac{\pi}{2}}x\ln(2)dx=\frac{{\pi}^{2}\ln^{2}(2)}{8}$
by using the identity $\displaystyle\sum_{k=1}^{\infty}\frac{x\cos(2kx)}{k}=-x\ln(\sin(x))-x\ln(2)$
and/or $\displaystyle \ln(\cos(x))=-\ln(2)-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos(2kx)}{k}$
I have used the first one to evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}}x\ln(\sin(x))dx$,  so I thought perhaps it could be used in some manner here.
I see some familiar things in the solution, but how to get there?.
Does anyone have any clever ideas?.
Thanks.
 [/INST] Here is an approach that avoids using the Fourier series expansions for $\ln (\cos x)$ and $\ln (\sin x)$.
Consider
$$I = \int_0^{\pi/2} x^2 \ln (\sin x) \ln (\cos x) \, dx.$$
Enforcing a substitution of $x \mapsto \pi/2 - x$ gives
\begin{align*}
I &= \int_0^{\pi/2} \left (\frac{\pi}{2} - x \right )^2 \ln \left [\sin \left (\frac{\pi}{2} - x \right ) \right ] \ln \left [\cos \left (\frac{\pi}{2} - x \right ) \right ] \, dx\\
&= \int_0^{\pi/2} \left (\frac{\pi}{2} - x \right )^2 \ln (\cos x) \ln (\sin x) \, dx\\
&= \frac{\pi^2}{4} \int_0^{\pi/2} \ln (\sin x) \ln (\cos x) \, dx - \pi \int_0^{\pi/2} x \ln (\sin x) \ln (\cos x) \, dx\\
& \qquad + \int_0^{\pi/2} x^2 \ln (\sin x) \ln (\cos x) \, dx,
\end{align*}
or
$$\int_0^{\pi/2} x \ln (\sin x) \ln (\cos x) \, dx = \frac{\pi}{4} \int_0^{\pi/2} \ln (\sin x) \ln (\cos x) \, dx.$$
The integral appearing on the right can be found by differentiating the beta function. As
$$\text{B}(x,y) = 2 \int_0^{\pi/2} \cos^{2x - 1} \theta \sin^{2y - 1} \theta \, d\theta,$$
we see that
$$\int_0^{\pi/2} \ln (\sin \theta) \ln (\cos \theta) \, d\theta = \frac{1}{8} \partial_x \partial_y \text{B}(x,y) \Big{|}_{x=y=1/2},$$
giving
$$\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx = \frac{\pi}{32} \partial_x \partial_y \text{B}(x,y) \Big{|}_{x=y=1/2}.$$
Since
$$\partial_x \text{B}(x,y) = \text{B}(x,y) [\psi(x) - \psi (x + y)] \quad \text{and} \quad \partial_y \text{B}(x,y) = \text{B}(x,y) [\psi(y) - \psi (x + y)],$$
where $\psi (z)$ is the digamma function, we have
$$\partial_x \partial_y \text{B}(x,y) = \text{B}(x,y) \left [ \left \{\psi(x) - \psi (x + y) \right \} \left \{\psi(y) - \psi (x + y) \right \} - \psi^{(1)}(x + y) \right ].$$
Thus
\begin{align*}
\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx &= \frac{\pi}{32} \text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) \left [ \left \{ \psi \left (\frac{1}{2} \right ) - \psi (1) \right \}^2 - \psi^{(1)}(1) \right ].
\end{align*}
Since
$$\text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) = \frac{\Gamma (1/2) \Gamma (1/2)}{\Gamma (1)} = \pi,$$
and
$$\psi^{(1)}(1) = \zeta (2) = \frac{\pi^2}{6} \quad \text{and} \quad \psi \left (\frac{1}{2} \right ) - \psi (1) = - 2\ln (2),$$
we have
$$\int_0^{\pi/2} x\ln (\sin x) \ln (\cos x) \, dx = \frac{\pi^2}{8} \ln^2 (2) - \frac{\pi^4}{192},$$
as expected.
</s>