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I'm stuck with the following integral $$I=\int_{-\infty}^{\infty}\frac{e^{-az^{2}}}{e^{-b-iz}-1}dz$$ It seems to me there exists a clever contour which encloses the pole at $z=ib$ while being somehow equivalent to the real line. Maybe someone has any ideas?

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    Looks like you want a semi-circle contour... – 2017-02-17
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    Actually, that's a lot of residues we'll have to sum over if we do a semi-circle contour. Note the poles occur every time $e^{-b-iz}=1$ – 2017-02-17
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    Semi-circular cannot work as the integrand diverges at say $z=i\infty$. And, you're right there are a lot of residues! – 2017-02-17
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    Aha! Do a rectangle with height of $2\pi$. :D – 2017-02-17
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    One method that will definitely work is: expand the denominator into a geometric series (assuming $b>0$, say) and integrate the resulting Gaussians. This will give a series as the answer. – 2017-02-18
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    In fact, I think this will give $I$ in terms of a $\theta$ function. – 2017-02-18
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    The thing is that this integral raised from the series you are talking about) It was something like $\sum_{n\geq0}\exp(-an^{2}-bn)$, then I used Hubbard-Stratonovich transformation $\exp(-an^{2})=\frac{1}{2\sqrt{\pi{a}}}\int_{\mathbb{R}}\exp\Big(-\frac{y^{2}}{4a}-iyn\Big)dy$, then summed the geometric series and here I am.) I think that I shall reject this approach and try to write the sum in terms of Jacobi Theta functions then& – 2017-02-19

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