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my question concerns the Mordell integral $$h(z;\tau):=\int_{-\infty}^\infty \frac{e^{\pi i\tau w^2-2\pi zw}}{\cosh(\pi w)}dw,\qquad \Im(\tau)>0,\quad z\in\mathbb{C},$$ which frequently occurs in the theory of mock modular forms.

I wonder whether one knows anything about the behaviour of this function for $\tau\rightarrow 0$. To be more precise, I am interested in the difference $$ h(3z-\tau;3\tau)-h(3z+\tau;3\tau)$$ where $z$ is real. One stumbles over this differnce when analyzing the modular transformation behaviour of the generating function of partition ranks.

Thanks a lot in advance!

MHMertens
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1 Answers1

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The Maple code

h := unapply(int(exp(I*Pi*tau*w^2-2*Pi*z*w)/cosh(Pi*w), w = -infinity .. infinity), z, tau):
series(h(3*z-tau, 3*tau)-h(3*z+tau, 3*tau), tau, 2)

produces $$\left(\int_{-\infty}^\infty \frac {e^{-6\pi z w}(3i\pi w^2+2\pi w)} {\cosh (\pi w)}\,dw- \int_{-\infty}^\infty \frac {e^{-6\pi z w}(3i\pi w^2-2\pi w)} {\cosh (\pi w)}\,dw\right)\tau +O(\tau^2)=$$ $$\tau \int_{-\infty}^\infty \frac {8\pi\, w\, e^{-6\pi w z+\pi w}}{e^{2\pi w}+1}\,dw+O(\tau^2),\, \tau \to 0.$$

user64494
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  • Does that last integral converge for arbitrary $z \in \mathbb{C}$? – S. Carnahan Apr 08 '14 at 16:38
  • The questioner assumes $z$ to be real. In this case the last integral converges for $|z|< \frac 1 3$ at least. – user64494 Apr 08 '14 at 16:48
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    I guess that's not bad, although $z$ really should be greater than $-1/6$. – S. Carnahan Apr 08 '14 at 16:57
  • There might be a tiny problem with this solution though: The Mordell integral as such is only defined for $\tau$ in the upper halfplane, in particular not in any neighbourhood of $0$. So a Taylor expansion around $0$ is not very well possible, is it? MAPLE probably doesn't care for that though... – MHMertens Apr 09 '14 at 08:11
  • BTW, MAPLE's limit command also tells me that the limit is $0$ which would be enough for my purpose, although I don't really trust MAPLE in this since I don't know how it computes these things. – MHMertens Apr 09 '14 at 08:44