Computing the Gaussian integral with Fourier methods?

Solution 1:

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

Then:

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

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

Then:

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

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:

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

Finally:

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

Solution 2:

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.

Solution 3:

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 $I = \sqrt{\pi}$.

Solution 4:

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.