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: $$\prod_{n=1}^\infty\left(1-e^{-24\!\;n\!\;x}\right)=\left(e^{-24\!\;x};\,e^{-24\!\;x}\right)_\infty\tag2$$
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.
One may just apply the Jacobi triple product to the Dedekind eta function, then perform a termwise integration that leads to a multiple of $\zeta(2)$.
