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In a previous question How to evaluate the Gauss sum using Mordell's Trick? I was trying to evaluate the integral

$$\int_\gamma \exp(2 \pi i z^2 / n)(\exp(2 \pi i z) + 1) dz$$

across the line $\gamma(t) = i t$ where $t \in [-R,R]$. $\gamma'(t) = i$ so I can change this from a contour integral to an integral across a range:

$$\int_{-R}^{R} i \exp(- 2 \pi i z^2 / n)(\exp(- 2 \pi z) + 1)$$

but it was said that this integral does not converge when $R$ tends to infinity. I am confused because I believe this integral should equal something else that does converge. Can anybody explain what I did wrong? How is it possible that this integral diverges when an equivalent integral converges?

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As the integrand is a continuous function, the integral from $-R$ to $R$ converges (as an ordinary Riemann integral). It can be expressed in terms of the error function. What might diverge is the limit as $R \to \infty$, i.e. the improper integral from $-\infty$ to $\infty$. Indeed if we take the part of the integral from $-R$ to $0$ and do a change of variables $z = -\sqrt{t}$, it becomes $$ i/2\int_{0}^{R^2}\!{\frac {{{\rm e}^{2\pi\sqrt {t}}}+1}{ \sqrt {t}}{{\rm e}^{{{-2\,i\pi t}/{n}}}}}\,{\rm d}t$$ and since $(\exp(2\pi \sqrt{t})+1)/\sqrt{t} \to \infty$ as $t \to \infty$, we see that it does indeed diverge.

Robert Israel
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