Closed form of the integral ${\large\int}_0^\infty e^{-x}\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)dx$

Question

While doing some numerical experiments, I discovered a curious integral that appears to have a simple closed form: $${\large\int}_0^\infty e^{-x}\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)dx\stackrel{\color{gray}?}=\frac{\pi^2}{6\sqrt3}\tag1$$ Could you suggest any ideas how to prove it?

The infinite product in the integrand can be written using q-Pochhammer symbol: $$\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)=\left(e^{-24\!\;x};\,e^{-24\!\;x}\right)_\infty\tag2$$

Answer 1

To be somewhat explicit. One may perform the change of variable, $q=e^{-x}$, $dq=-e^{-x}dx$, giving $$ {\large\int}_0^\infty e^{-x}\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)dx={\large\int}_0^1 \prod_{n=1}^\infty\left(1-q^{24n}\right)dq\tag1 $$ then use the identity (the Euler pentagonal number theorem) $$ \prod_{n=1}^\infty\left(1-q^{n}\right)=\sum_{-\infty}^{\infty}(-1)^nq^{\large \frac{3n^2-n}2} $$ to get

$$ {\large\int}_0^\infty e^{-x}\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)dx=\sum_{-\infty}^{\infty}{\large\int}_0^1(-1)^n q^{12 (3n^2-n)}dq=\sum_{-\infty}^{\infty}\frac{(-1)^n}{(6n-1)^2}=\frac{\pi ^2}{6 \sqrt{3}} $$

The last equality is obtained by converting the series in terms of the Hurwitz zeta function and by using the multiplication theorem.

Answer 2

One may just apply the Jacobi triple product to the Dedekind eta function, then perform a termwise integration that leads to a multiple of $\zeta(2)$.