Integral formula for $\int_{0}^{\infty}e^{-3\pi x^{2}}((\sinh \pi x)/(\sinh 3\pi x))\,dx$ by Ramanujan

Towards the end of G. N. Watson’s (one of the joint authors of famous book “A Course of Modern Analysis”) paper “The Final Problem: An Account of the Mock Theta Functions” the following formula of Ramanujan is mentioned: $$\int_{0}^{\infty}e^{-3\pi x^{2}}\frac{\sinh \pi x}{\sinh 3\pi x}\,dx = \frac{1}{e^{2\pi/3}\sqrt{3}}\sum_{n = 0}^{\infty}\frac{e^{-2n(n + 1)\pi}}{(1 + e^{-\pi})^{2}(1 + e^{-3\pi})^{2}\dots(1 + e^{-(2n + 1)\pi})^{2}}\tag{1}$$ where the term corresponding to $n = 0$ in the sum on the right is $1$.

Is there way to establish this exotic integral formula? Or a reference to any existing proof of $(1)$ would be of great help.

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I re-posted this on MO and got the desired answer. The answer to the question is contained in the same paper of G. N. Watson which is referred to in the question.

The integral in the question comes up in the transformation formulas for the one of the several mock theta functions defined by Ramanujan. The series in equation $(1)$ of the question is the value of a certain mock theta function $\omega(q)$ at the point $q = -e^{-\pi}$.

Let $q$ be real with $|q| < 1$ and we define the mock theta function $\omega(q)$ via the equation $$\omega(q) = \sum_{n = 0}^{\infty}\frac{q^{2n(n + 1)}}{(1 – q)^{2}(1 – q^{3})^{2}\dots (1 – q^{2n + 1})^{2}}\tag{1}$$ so that $$\omega(-q) = \sum_{n = 0}^{\infty}\frac{q^{2n(n + 1)}}{(1 + q)^{2}(1 + q^{3})^{2}\dots (1 + q^{2n + 1})^{2}}\tag{2}$$ and the question asks us to prove $$\int_{0}^{\infty}e^{-3\pi x^{2}}\frac{\sinh \pi x}{\sinh 3\pi x}\,dx = \frac{q^{2/3}}{\sqrt{3}}\omega(-q)\tag{3}$$ with $q = e^{-\pi}$.

Watson proves a transformation formula for $\omega(-q)$ in his paper which uses the integral mentioned in the question. He shows that if $\alpha, \beta$ are positive real numbers such that $\alpha\beta = \pi^{2}$ and $q = e^{-\alpha}, q_{1} = e^{-\beta}$ then $$q^{2/3}\omega(-q) + \sqrt{\frac{\pi}{\alpha}}q_{1}^{2/3}\omega(-q_{1}) = 2\sqrt{\frac{3\alpha}{\pi}}I(\alpha)\tag{4}$$ where $$I(\alpha) = \int_{0}^{\infty}e^{-3\alpha x^{2}}\frac{\sinh \alpha x}{\sinh 3\alpha x}\,dx\tag{5}$$ Putting $\alpha = \beta = \pi$ in $(4)$ and noting that $q = q_{1} = e^{-\pi}$ we get equation $(3)$. So the crux of the problem is to prove the transformation formula $(4)$ and this is a difficult task which Watson achieved via finding another suitable series representation for $\omega(q)$ and using residue calculus to convert the series for $\omega(q)$ into a contour integral. See Watson’s paper for more details.