The following is a re-post from MSE because I did not get any answer even after offering a bounty.

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: \begin{align}&\int_{0}^{\infty}e^{-3\pi x^{2}}\frac{\sinh \pi x}{\sinh 3\pi x}\,dx\notag\\ &\,\,\,\,\,\,\,\,= \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} \end{align} 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.


1 Answer 1


This relation seems to be part of the story of Ramanujan's mock theta function. The identity you discuss is a specialization of transformation properties of a mock theta function evaluated at $\tau =i$. See this paper of Zwegers for proofs of the general transformation identity (and he gives other references to earlier work of Watson): in particular, look at section 3 and Lemma 3.2 of Zwegers's paper. Other surveys that you may find of interest are Ono and Duke.

  • 1
    $\begingroup$ Thanks. I now get the full picture. This integral is related to $\omega(-q)$ of Watson's paper. I had not read those transformations formulas (just read proofs of algebraic identities between mock theta functions) otherwise the answer to my question would have been obvious. $\endgroup$ Commented Aug 1, 2015 at 7:43

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