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I was teaching myself complex analysis using Gamelin's book Complex analysis. In the chapter about Cauchy's theorem, he presents the next problem:

By integrating $e^{-z^{2}/2}$ around a rectangle with vertices $\pm R$, $it\pm R$, and sending $R$ to $\infty$, show that $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-x^{2}/2}e^{-itx}dx = e^{-t^{2}/2}, \hspace{0.6cm} -\infty < t < \infty. $$ Use the known value of the integral for $t = 0$. $Remark$. This shows that $e^{-x^{2}/2}$ is an eigenfunction of the Fourier transform with eigenvalue 1.

I was trying to use Cauchy's theorem for get $\int e^{-z^{2}/2}$ around the rectangle is zero, and from that, decompose it into its real and imaginary parts, but I'm getting stuck. Should I change the approach of the problem?

MrJordanCs
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  • The main astute idea is to write your expression under the equivalent form $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{t^{2}/2}e^{-x^{2}/2}e^{-itx}dx = 1$, otherwise said $\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{\tfrac12 (t-ix)^2}dx = 1$. – Jean Marie Jul 01 '22 at 05:53
  • I advise you to learn also on Needham's "Visual complex analysis". But there are other alternative books : see here – Jean Marie Jul 01 '22 at 05:57
  • See this answer here – Jean Marie Jul 01 '22 at 08:43

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