How to calculate $ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx $? How to calculate $$ \int_0^{\pi/2}\log(1+\sin(x))\log(\cos(x)) \,dx \,\,?$$
I tried to use the Fourier series of log sine and log cos and I got that the integral is equal to :
$$ \frac{\pi^2}{24}-\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{n+k}}{k(4k^2-(2n-1)^2)}$$
has anyone a idea to how to find the closed-form of the last series or how to start out differently with the integral?
 A: Substitute $t=\tan\frac x2$
\begin{align} &\int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx \\
=&\>2\int_0^{1}\frac{\ln\frac{(1+t)^2}{1+t^2}\ln \frac{1-t^2}{1+t^2} }{1+t^2}\,dt
=\>4I_1 +4 I_2 -2I_3- 6I_4+2I_5
\end{align}
where, per the results
\begin{align}
I_1 &= \int_0^1 \frac{\ln (1+t)\ln(1-t)}{1+t^2} dt
= -G \ln 2-K+\frac{3 \pi ^3}{128}+\frac{3\pi}{32} \ln ^22\\
 I_2 &= \int_0^1 \frac{\ln^2(1+t)}{1+t^2} dt
= -2 G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{16} \ln ^22 \\
I_3 &= \int_0^1 \frac{\ln (1+t^2)\ln(1-t)}{1+t^2} dt
=  -\frac{1}{2} G \ln 2+4 K -\frac{5 \pi ^3}{64}+\frac{\pi}{8}  \ln ^22 \\
 I_4 &= \int_0^1 \frac{\ln (1+t^2)\ln(1+t)}{1+t^2} dt
=  -\frac{5}{2} G \ln 2-4 K+\frac{7 \pi ^3}{64}+\frac{3\pi}{8} \ln ^22\\
 I_5 &= \int_0^1 \frac{\ln^2(1+t^2)}{1+t^2} dt
= -2 G \ln 2+4 K-\frac{7 \pi ^3}{96}+\frac{7\pi}{8} \ln ^22
\end{align}
with $K= \Im\text{Li}_3\left(\frac{1+i}{2}\right)$. Together
$$ \int_0^{\pi/2}\ln(1+\sin x)\ln(\cos x) \,dx 
=4\Im\text{Li}_3\left(\frac{1+i}{2}\right)-\frac{11\pi^3}{96}+\frac{3\pi}8\ln^22
$$
A: We have a closed form for the inner sum
$$S_k=\sum_{n=1}^{\infty}\frac{(-1)^{n+k}}{k(4k^2-(2n-1)^2)}$$
$$S_k=(-1)^k \frac{\Phi \left(-1,1,\frac{1}{2}-k\right)-\Phi
   \left(-1,1,\frac{1}{2}+k\right)}{8 k^2}$$ where appears the Hurwitz-Lerch transcendent function. This can rewrite
$$S_k=(-1)^k \frac{-\psi \left(\frac{1}{4}-\frac{k}{2}\right)+\psi
   \left(\frac{3}{4}-\frac{k}{2}\right)+\psi
   \left(\frac{k}{2}+\frac{1}{4}\right)-\psi
   \left(\frac{k}{2}+\frac{3}{4}\right) }{16 k^2 }$$
