Let $\xi \in \mathbb{R}$. Use Cauchy's Theorem to prove that $\int_{-\infty}^{\infty}e^{-\pi x^2}e^{-2\pi ix \xi} dx = e^{-\pi \xi^2}$.

Question

It wants us to prove that $\int_{-\infty}^{\infty}e^{-\pi x^2}e^{-2\pi ix \xi} dx = e^{-\pi \xi^2}$ by integrating $f(z) = e^{-\pi z^2}$ over the rectangle with vertices $\pm n, \pm n + i \xi$ and taking the limit $n \to \infty$. I have tried to do this using Cauchy's Integral Formula, but it ends up with a nasty integral that results in $\text{erf}(z)$, which I know for a fact we cannot use. Any help on how to start with this problem? I have seen similar questions but none that asked to find it by integrating over this $2n \times i \xi$ rectangle.

Answer 1

We first estimate the integral of $e^{-\pi z^2}$ over $\left[\pm n,\pm n+i\xi\right]$. Note that for $z=\pm n+ic$ on the segment, we have $$\left | e^ {z^2} \right | =e^{c^2-\pi n^2 }\leqslant e^{\xi^2-\pi n^2}, $$ therefore $$\left | \int_{\pm n}^{\pm n+i\xi } e^{-\pi z^2}\mathrm d z \right | \leqslant \left | \xi \right |e^{\xi^2-\pi n^2}\to 0. $$ Now fix $n$, it is easy to see that the integrand is analytic.Recall that $$\text{The Rectangle}=\left [ -n,n \right ] \cup \left [ -n+i\xi,n +i\xi \right ] \cup \left [ n,n+i\xi \right ] \cup \left [ -n,-n+i\xi \right ] $$ and the whole integral over the rectangle is $0$. Take $n\to \infty$ and we see that $$\int_{-\infty+i\xi }^{\infty+i\xi} e^{-\pi z^2}\mathrm d z=\int_{-\infty }^{\infty} e^{-\pi z^2}\mathrm d z.$$ Therefore we have $$\int_{-\infty }^{\infty} e^{-\pi \left ( z+i\xi \right ) ^2}\mathrm d z=\int_{-\infty }^{\infty} e^{-\pi z^2}\mathrm d z=1.$$ Take the $e^{\pi\xi^2}$ out from the left hand side and we are done.

Answer 2

By Cauchy's theorem, we have; $$\oint f(z)dz=\oint e^{-\pi z^2}dz=0$$

$$\therefore\int_{-n}^{n}e^{-\pi z^2}dz+\int_{n}^{n+i\zeta}e^{-\pi z^2}dz+\int_{n+i\zeta}^{-n+i\zeta}e^{-\pi z^2}dz+\int_{-n+i\zeta}^{-n}e^{-\pi z^2}dz=0$$

In the limit as $n\rightarrow\infty$ we will have;

$$0=\int_{-\infty}^{\infty}e^{-\pi z^2}dz+\lim_{n\rightarrow\infty}\int_{n+i\zeta}^{-n+i\zeta}e^{-\pi z^2}dz$$

Applying the substitution $z=m+i\zeta$ yields;

$$\int_{-\infty}^{\infty}e^{-\pi z^2}dz=-\lim_{n\rightarrow\infty}\int_{n}^{-n}e^{-\pi (m+i\zeta)^2}dm=\int_{-\infty}^{\infty}e^{-\pi (m+i\zeta)^2}dm$$

Utilizing the fact that $\small\int_{-\infty}^{\infty}e^{-\pi x^2}dx=1$ then gives us;

$$1=e^{\pi\zeta^2}\int_{-\infty}^{\infty}e^{-\pi m^2}e^{-2\pi i m \zeta}dm$$

Leaving us with the final identity;

$$\int_{-\infty}^{\infty}e^{-\pi m^2}e^{-2\pi i m \zeta}dm=e^{-\pi\zeta^2}$$

as expected.