Is there a closed-form for $\sum _{k=1}^{\infty }\frac{\operatorname{Si}\left(k\right)}{k^2}$? So far I've got this:
$$\sum _{k=1}^{\infty }\frac{\operatorname{Si}\left(k\right)}{k^2}=\int _0^1\left(\sum _{k=1}^{\infty }\frac{\sin \left(kx\right)}{k^2}\right)\frac{1}{x}\:dx$$
$$=\int _0^1\frac{\operatorname{Cl}_2\left(x\right)}{x}\:dx=\operatorname{\mathfrak{R}} \left\{\int _0^1\ln \left(1-e^{ix}\right)\ln \left(x\right)\:dx\right\}$$
$$=\frac{1}{2}\ln \left(2\right)\int _0^1\ln \left(x\right)\:dx+\frac{1}{2}\int _0^1\ln \left(1-\cos \left(x\right)\right)\ln \left(x\right)\:dx$$
$$=-\ln \left(2\right)+2\ln \left(2\right)\int _0^{\frac{1}{2}}\ln \left(\sin \left(x\right)\right)\:dx+2\int _0^{\frac{1}{2}}\ln \left(\sin \left(x\right)\right)\ln \left(x\right)\:dx$$
Where $\operatorname{Si}\left(k\right)=\int _0^k\frac{\sin \left(x\right)}{x}\:dx$ is the sine integral and $\operatorname{Cl}_2\left(x\right)=\sum _{k=1}^{\infty }\frac{\sin \left(kx\right)}{k^2}$ is the Clausen function.
Note that:
$$\int _0^{\frac{1}{2}}\ln \left(\sin \left(x\right)\right)\:dx=\frac{i}{8}-\frac{i}{2}\zeta \left(2\right)-\frac{1}{2}\ln \left(1-e^i\right)+\frac{i}{2}\operatorname{Li}_2\left(e^i\right)+\frac{1}{2}\ln \left(\sin \left(\frac{1}{2}\right)\right)$$
But I've no idea how to even start with that other integral, how to proceed with it or the sum?
 A: By partial integration
$$
\int_0^u \ln (\sin x)\ln x dx
=
\ln (\sin x) x (\ln x-1)\mid_{x=0}^u -\int _0^u \cot x ( x \ln x -x) dx.
$$
So one needs first
$$
\int_0^u x \cot x dx =
\int_0^u x / \tan x dx =
\operatorname{Cl}_2(u)+\operatorname{Cl}_2(u+\pi) -i\frac{\pi^2}{4}+u\left(\ln(1-e^{2iu}-iu/2)+\frac{i}{2}(\pi u -u^2+\pi^2)
\right)$$
taken from eq (94) of Coffey's "On a three-dimensional symmetric Ising tetrahedron.." J. Math. Phys. 49 (2008) 043510
doi:10.1063/1.2902996 .
And second one needs
$$
\int_0^u x \cot x \ln x dx
$$
which I can only imagine to handle via the Taylor series
$$
x\cot x = 1 - \frac{1}{3}x^2-\frac{1}{45}x^4-\frac{2}{945}x^6-\cdots
$$
and then taking each term $\int x^j \ln x dx$  separately. This converges quickly
$$ \int_0^{1/2} x \cot x \ln x dx = [-\frac12 \log(2)-\frac12] + [\frac{1}{72}\log (2)+\frac{1}{216}]+[\frac{1}{7200}\log(2)+\frac{1}{36000}] +\cdots \approx -0.8321908558687$$
where the square brackets indicate the contributions from $j=0,2,4,\ldots$.
