Find $\int_{0}^{\frac{\pi}{2}} \ln(\sin(x)) \ln( \cos(x))\,\mathrm dx$ $$\large\int \limits_{0}^{\frac{\pi}{2}} \ln(\sin(x)) \ln( \cos(x))\mathrm dx$$
TL;DR: Is there an elegant way of integrating this? I've reduced it to a series, detailed below, but the closed form eludes me, and the only solution I've seen uses a rabbit-out-of-the-hat approach.
For the series,
$$\begin{align} \int \limits_{0}^{\frac{\pi}{2}} \ln(\sin(x)) \ln( \cos(x))\,\mathrm dx&=-\sum \limits_{k=1}^\infty \frac{1}{k}\int \limits_{0}^{\frac{\pi}{2}} \sin^{2k}(x)\ln(\sin(x))\,\mathrm dx\\
&=-\sum \limits_{k=1}^\infty \frac{1}{k}\int_{0}^{\frac{\pi}{2}} \sin^{2k}(x)\ln(\sin(x))\,\mathrm dx\\
&=-\sum \limits_{k=1}^\infty \frac{1}{k}\int_{0}^{1} u^{2k}\ln(u)\frac{\mathrm du}{1-u^2}\\
&=-\sum \limits_{k=1}^\infty \frac{1}{k} \left(\int_{0}^{1} u^{2k}\ln(u)\,\mathrm du+ \sum \limits_{j=1}^\infty \frac{(2j-1)!!}{(2j)!!} \int_{0}^{1} u^{2(k+j)}\ln(u)\,\mathrm du \right)\\
&=\sum \limits_{k=1}^\infty \frac{1}{k} \left( \frac{1}{(2k+1)^2}+\sum \limits_{j=0}^\infty  \frac{(2j-1)!!}{(2j)!!} \frac{1}{(2(k+j)+1)^2} \right)\end{align}$$
, which converges! Both summations are sort-of-justified as $|\sin(x)|, |u|\le1$,with equality only reached at one of the limits of integration, not in between.
 A: $$ \beta(x,y) = 2\int_0`^{\frac{\pi}{2}} \sin^{2x-1}(t) \cos^{2y-1}(t) \ dt $$
$$ \frac{\partial }{\partial x} \beta(x,y) = 4\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t)\ln(\sin t) \cos^{2y-1}(t) \ dt$$
$$ \frac{\partial}{\partial y} \left( \frac{\partial}{\partial x} \beta(x,y) \right) = 8\int_0^{\frac{\pi}{2}} \sin^{2x-1}(t) \ln(\sin t ) \ln(\cos t) \cos^{2y-1}(t) \ dt $$
and we have $$ \beta(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} $$
so differentiate and put $ x =\frac{1}{2} , y = \frac{1}{2} $
$$ \psi \left(\frac{1}{2} \right) = -2\ln 2 - \gamma $$
$$ \psi(1) = -\gamma $$
$$ \psi^{(1)}(1) = \frac{\pi^2}{6} $$
$$ \beta \left( \frac{1}{2} , \frac{1}{2} \right) = \frac{\Gamma \left( \frac{1}{2} \right)^2}{\Gamma(1)} = \pi $$ 
thus you'll have the integral $ = \frac{\pi}{8} \left(4\ln(2)^2 - \frac{\pi^2}{6} \right) $
A: Use the Fourier series
$$\ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k \\
  \ln(\cos(x)) = -\ln(2) - \sum_{k=1}^\infty (-1)^k \frac{\cos(2kx)}k$$
and expand the integrand to
$$\begin{align*}
\ln(\sin(x)) \ln(\cos(x)) &= \ln^2(2) + \ln(2) \sum_{k=1}^\infty \frac{1+(-1)^k}2 \frac{\cos(2kx)}k \\[1ex]
& \qquad+ \left(\sum_{k=1}^\infty \frac{\cos(2kx)}k\right) \left(\sum_{\ell=1}^\infty (-1)^\ell \frac{\cos(2\ell x)}\ell\right) \\[2ex]
&= \ln^2(2) + \ln(2) \sum_{k=1}^\infty \frac{1+(-1)^k}2 \frac{\cos(2kx)}x \\[1ex]
& \qquad + \sum_{k=1}^\infty \frac{(-1)^k \cos^2(2kx)}{k^2} + 2 \sum_{k\neq\ell} \frac{(-1)^\ell \cos(2kx) \cos(2\ell x)}{k\ell}
\end{align*}$$
Now integrate. We have for all non-negative integers $k,\ell$
$$\int_0^{\pi/2} \cos(2kx)\cos(2\ell x) \, dx = \begin{cases}\dfrac\pi2 & \text{if }k=\ell=0 \\ \dfrac\pi4 & \text{if }k=\ell>0 \\ 0 & \text{if }k\neq\ell\end{cases}$$
so upon swapping integrals with sums, we find
$$\begin{align*}
\int_0^{\pi/2} \ln(\sin(x))\ln(\cos(x)) \, dx &= \frac\pi2 \ln^2(2) + (0\times\ln(2)) + \frac\pi4 \sum_{k=1}^\infty \frac{(-1)^k}{k^2} + (2\times0) \\[1ex]
&= \boxed{\frac\pi2 \ln^2(2) - \frac{\pi^3}{48}}
\end{align*}$$
