How to calculate $\int_0^\infty e^{-x} \prod_{n=1}^\infty (1-e^{-24nx} ) dx$

Question

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.

Answer 1

$$e^{-x}\prod_{n=0}^\infty(1-e^{-24nx}) =\sum_{k=-\infty}^\infty(-1)^ke^{-(6k+1)^2x}$$ 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.

Answer 2

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 $$ \psi'(s)+\psi'(1-s) = \frac{\pi^2}{\sin^2(\pi s)}.$$