There are many proofs that

$$\int_{-\infty}^\infty e^{-x^2} \, \mathrm dx = \sqrt{\pi}.$$

For example, using a change to polar coordinates, differentiation under the integral sign, and the theory of the Gamma function. However, I am told there are very natural and simple ways to evaluate it using methods from Fourier analysis. This is not particularly surprising to me, considering, for example, that the Gaussian is its own Fourier transform, but I haven't seen an actual proof. So, how does one compute the Gaussian integral using methods from Fourier analysis?


$$F(\omega) = \frac{1}{2 \pi} \int_{-\infty}^{+\infty} \exp(\frac{-t^2}{2}) \exp(- i \omega t) dt$$


$$F(\omega) = \frac{1}{\pi} \int_{0}^{+\infty} \exp(\frac{-t^2}{2}) \cos( \omega t) dt$$

Justify the derivation and integrate by parts to show that: $$F'(\omega) = - \omega F(\omega)$$


$$F(\omega) = C \exp(\frac{-\omega^2}{2})$$

Yet we know the Fourier inverse:

$$\exp(\frac{-x^2}{2}) = \int_{-\infty}^{+\infty} F(\omega) \exp( i \omega x) d\omega$$

Replace the formula for $F(\omega)$, we get almost the first formula we used, thus:

$$C = \frac{1}{\sqrt{2\pi}}$$

Take $\omega=0$ to conclude:

$$F(0) = C = \frac{1}{2 \pi} \int_{-\infty}^{+\infty} \exp(\frac{-t^2}{2}) dt$$


$$\sqrt{2} \int_{-\infty}^{+\infty} \exp(-t^2) dt = \sqrt{2\pi}$$


A way is to make use of the Poisson summation formula. I will work with the Fourier transform $$\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) \exp(-2 \pi i \xi x) dx$$ The Poisson summation formula states that $$\sum_{\xi \in \mathbb{Z}} \hat{f}(\xi) = \sum_{n \in \mathbb{Z}} f(n).$$ Now take $f(x) = \exp(-\pi x^2)$. We then get that \begin{align} \hat{f}(\xi) & = \int_{-\infty}^{\infty} \exp(- \pi x^2) \exp(-2 \pi i \xi x) dx\\ & = \int_{-\infty}^{\infty} \exp(-\pi(x+i \xi)^2 - \pi \xi^2) dx\\ & = \exp( - \pi \xi^2)\int_{-\infty}^{\infty} \exp(-\pi(x+i \xi)^2) dx\\ & = \exp( - \pi \xi^2)\int_{-\infty + ic}^{\infty+ic} \exp(-\pi x^2) dx \end{align} By integrating from $-\infty+ic$ to $\infty + ic$, I mean integrate along the line Im$(x) = c$ from left to right. Now since the integrand is analytic, we can move this contour to $X$ axis and conclude that $$\int_{-\infty + ic}^{\infty+ic} \exp(-\pi x^2) dx = \int_{-\infty}^{\infty} \exp(-\pi x^2) dx$$ Hence, we get that $$\hat{f}(\xi) = C \exp( - \pi \xi^2)$$where $C = \displaystyle \int_{-\infty}^{\infty} \exp(-\pi x^2) dx$. Now make use of the Poisson summation formula to get that $$C \left(\sum_{\xi \in \mathbb{Z}} \exp(-\pi \xi^2) \right) = \left(\sum_{x \in \mathbb{Z}} \exp(-\pi x^2) \right)$$ We can afford to cancel $\displaystyle \left(\sum_{x \in \mathbb{Z}} \exp(-\pi x^2) \right)$ since it converges and hence we can conclude that $$C = \displaystyle \int_{-\infty}^{\infty} \exp(-\pi x^2) dx = 1$$ Suitable scaling gives you the integral and answer you are looking for.


Set $f(x) = e^{-x^2}$ and $I = \int_{\mathbb{R}} e^{-x^2} \, dx$. Then compute the Fourier transform of $f$

$$\widehat{f}(y) = \int_{\mathbb{R}} e^{-x^2-2i\pi x y} \, dx = e^{-(\pi y)^2} \int_{\mathbb{R}} e^{-(x+i \pi y)^2} \, dx$$

Using the residue theorem, we can show the value of $\int_{\mathbb{R}} e^{-(x+i \pi y)^2} \, dx$ does not depend on $y$ (roughly, integrate $f$ on an horizontal rectangle whose width goes to infinity). So $\widehat{f}(y) = I . f(\pi y)$. Now use Fourier inversion formula

$$f(x) = I . \int_{\mathbb{R}} f(\pi y) e^{-2i\pi x y} \, dy \underset{t = \pi y}{=} \frac{I}{\pi} . \int_{\mathbb{R}} f(t) e^{-2i x t} \, dt = \frac{I}{\pi} \widehat{f}\left(\frac{x}{\pi}\right) = \frac{I^2}{\pi} f(x)$$

Since $I \ge 0$ (because $f \ge 0$), you get $I = \sqrt{\pi}$.


In this paper http://www.math.uconn.edu/~kconrad/blurbs/analysis/gaussianintegral.pdf you have as many as 10 proofs of this result, the 10th being by means of Fourier Transform.


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