I have been working on the following problem from Gamelin VII.1 problem 6.

Consider the integral $$ J = \int_{\partial D_R} \frac{\exp\bigr( \pi i (z - 1/2)^2 \bigl)}{1 - \exp(-2\pi i z)} \mathrm{d}z $$
where $D_R$ is the parallelogram with $\pm \frac12 \pm (1+i) R$.

a) Use the residue theorem to shot that the integral is $(1+i)/\sqrt2$.

b) By parameterizing the sides of the parallelogram, show that the integral tends to $$ (1+i) \int_{-\infty}^\infty e^{-2 \pi t^2} > \mathrm{d}t $$ as $R \to \infty$.

c) Use a) and c) to show that $$ \int_{-\infty}^{\infty} e^{-x^2} \,\mathrm{d}s = \sqrt{\pi} $$

I have been able to solve a) using the residue theorem. No matter what $R$ is, the only singularity will be located at $z=0$. So

The contour integral

$$ J = 2\pi i \cdot \text{Res}\left[ \frac{\exp\bigr( \pi i (z - 1/2)^2 \bigl)}{1 - \exp(-2\pi i z)} , z=0\right] = 2\pi i \left[ \frac{\exp\bigr( \pi i (0 - 1/2)^2 \bigl)}{2 i \pi \exp(-2\pi i \cdot 0)} \right] = e^{\pi i/4} $$

Which is the same as $(1+i)/\sqrt{2}$.

Now my problem lies in showing b). In particular how do I prove that the integrals over $\gamma_4$ and $\gamma_2$ tends to zero as $R \to \infty$?


1 Answer 1


Note that on these segments, $\arg(z)$ gets closer to $\pi/4$ and $5\pi/4$. Thus, the $\arg(z^2)$ approaches $\pi/2$ in both cases. Thus, $$ \pi i(z-1/2)^2\text{ approaches }-\pi|z|^2\sim-2\pi R^2 $$ The denominator stays away from $\mathbb{R}$ which is the only place that the it can vanish. That is, on the segment in the first quadrant, the denominator has absolute value about $e^{2\pi R}$ and in the third quadrant, the denominator has absolute value approaching $1$.

  • $\begingroup$ Could you say that along the top of the parallelogram, $$ \text{max} \ \Big|\exp \left(\pi i \left( z-1/2\right)^{2} \right) \Big|= \text{max} \ \Big| \text{exp} \left( \pi i \left( t+R+iR - 1/2 \right)^{2} \right)\Big| $$ $$ = \text{max} \ e^{-2 \pi R^{2}} e^{-2 \pi Rt} e^{\pi R} = e^{-2 \pi R^{2}} e^{2 \pi R}$$ And $$ \text{min} \ |1-e^{-2 \pi i z}| = e^{2 \pi R }-1$$ Then the magnitude of the integral along the top of the parallelogram is bounded from above by $$ 1 \times \frac{e^{-2 \pi R^{2}}}{1-e^{-2 \pi R}} \to 0 \ \text{as} \ R \to \infty \ ?$$ $\endgroup$ Mar 10, 2014 at 19:13
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    $\begingroup$ @RandomVariable: the estimate would go something like that. I think the first estimate would be $e^{-2\pi R^2+\pi R}$, but the difference is irrelevant to the answer. $\endgroup$
    – robjohn
    Mar 10, 2014 at 19:53
  • $\begingroup$ The reason I have $e^{2 \pi R}$ is because on the interval $-\frac{1}{2} \le t \le \frac{1}{2}$, $e^{-2 \pi R t}$ is maximized when $t = -\frac{1}{2}$. Then $\text{max} \ e^{-2 \pi R^{2}} e^{-2 \pi Rt} e^{\pi R} = e^{-2 \pi Rt} e^{\pi R} e^{\pi R} = e^{-2 \pi R^{2}} e^{2 \pi R} $. $\endgroup$ Mar 10, 2014 at 20:15
  • $\begingroup$ $$\begin{align}\mathrm{Re}\left(\pi i\left(t-\tfrac12+R(1+i)\right)^2\right) &=\mathrm{Re}\left(-2\pi R^2+2\pi iR(1+i)\left(t-\tfrac12\right)+\pi i\left(t-\tfrac12\right)^2\right)\\ &=-2\pi R^2-2\pi R\left(t-\tfrac12\right)\end{align}$$ which is greatest when $t=0$: $-2\pi R^2+\pi R$ $\endgroup$
    – robjohn
    Mar 10, 2014 at 20:29
  • $\begingroup$ The variable $t$ is varying from $- \frac{1}{2}$ to $\frac{1}{2}$, right? $-\frac{1}{2}$ And when $t=-\frac{1}{2}$, we have $\exp( -2 \pi R^{2} + 2 \pi R)$. That's larger than $\exp(-2 \pi R^{2} + \pi R)$, isn't it? $\endgroup$ Mar 10, 2014 at 21:25

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