How to calculate $\int_0^\infty e^{-x} \prod_{n=1}^\infty (1-e^{-24nx} ) dx$ How to calculate  $$\int_0^\infty e^{-x} \prod_{n=1}^\infty (1-e^{-24nx} ) dx$$  I'm stuck with the integral as I don't know how to handle the product. 
 A: As a followup to Lord Shark's answer, we have
$$ \sum_{k\geq 0}\left[\frac{1}{(6k+1)^2}-\frac{1}{(6k+5)^2}\right]=-\sum_{n\geq 0}\int_{0}^{1} \left(x^{6k}-x^{6k+4}\right)\log(x)\,dx\\=-\int_{0}^{1}\frac{1-x^4}{1-x^6}\log(x)\,dx $$
and by partial fraction decomposition and the digamma machinery this equals
$$ \tfrac{1}{72}\left[-\psi'\left(\tfrac{1}{6}\right)-5\,\psi'\left(\tfrac{1}{3}\right)+5\,\psi'\left(\tfrac{2}{3}\right)+\psi'\left(\tfrac{5}{6}\right)\right]$$
which does not simplify much further due to the unlucky pattern of signs. On the other hand
$$ \sum_{k\geq 0}\frac{(-1)^k}{(6k+1)^2}=\int_{0}^{1}\frac{-\log x}{1+x^6}\,dx =\tfrac{1}{144}\left[\psi'\left(\tfrac{1}{12}\right)-\psi'\left(\tfrac{7}{12}\right)\right]$$
by the same principle, and 
$$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(6k+1)^2}=\tfrac{1}{144}\left[\psi'\left(\tfrac{1}{12}\right)-\psi'\left(\tfrac{5}{12}\right)-\psi'\left(\tfrac{7}{12}\right)+\psi'\left(\tfrac{11}{12}\right)\right] $$
does simplify into
$$ \sum_{k\in\mathbb{Z}}\frac{(-1)^k}{(6k+1)^2} = \frac{\pi^2}{6\sqrt{3}}$$
due to the reflection formula for the trigamma function
$$ \psi'(s)+\psi'(1-s) = \frac{\pi^2}{\sin^2(\pi s)}.$$
A: $$e^{-x}\prod_{n=0}^\infty(1-e^{-24nx})
=\sum_{k=-\infty}^\infty(-1)^ke^{-(6k+1)^2x}$$
by Euler's pentagonal number formula. Integrating termwise gives
$$\sum_{k=-\infty}^\infty\frac{(-1)^k}{(6k+1)^2}
=\sum_{m=1}^\infty\frac{\chi(m)}{m^2}$$
where $\chi$ is the Dirichlet character modulo $6$ with $\chi(1)=1$
and $\chi(-1)=-1$.  Alas I believe this is an L-series evaluation
which doesn't have a simple closed form.
