# [Math] Gaussian Integral

integration

Consider the following Gaussian Integral $$I = \int_{-\infty}^{\infty} e^{-x^2} \ dx$$

The usual trick to calculate this is to consider $$I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} \ dx \right) \left(\int_{-\infty}^{\infty} e^{-y^{2}} \ dy \right)$$

and convert to polar coordinates. We get $\sqrt{\pi}$ as the answer.

Is it possible to get the same answer by considering $I^{3}, I^{4}, \dots, I^{n}$?

Define $$I_n = \prod_{i=1}^n \int^{\infty}_{-\infty} e^{-x_i^2}\,dx_i = \int_{\mathbb{R}^n} e^{-|x|^2}\,dx,$$ where $x = (x_1,\ldots,x_n)$ and $|x| = \sqrt{x_1^2 + \cdots + x_n^2}$.
$$I_n = \int_0^\infty\left\{\int_{\partial B(0,r)} e^{-|x|^2}\,dS\right\}\,dr.$$ For $e^{-|x|^2} = e^{-r^2}$ on $\partial B(0,r)$ which is a constant when the radius is fixed, hence above integral reads: $$I_n = \int_0^\infty e^{-r^2}\left\{\int_{\partial B(0,r)} 1\,dS\right\}\,dr.$$ Now $$\int_{\partial B(0,r)} 1\,dS = |\partial B(0,r)|= \omega_n r^{n-1},$$ which is the surface area of the $(n-1)$-sphere with radius $r$, and $\omega_n$ is the surface area of the unit $(n-1)$-sphere: $$\omega_n = \frac{n\pi^{n/2}}{\Gamma\left(1+\frac{n}{2}\right)},$$ notice this can be computed by a recursive relation or taking derivative of the volume element of the $n$-ball.
Hence $$I_n = \omega_n\int_0^\infty e^{-r^2}r^{n-1}\,dr = \frac{1}{2}\omega_n \Gamma\left(\frac{n}{2}\right) .$$
Now by Gamma function's property (proved by integration by parts): $$\Gamma\left(1+\frac{n}{2}\right) = \frac{n}{2}\Gamma\left(\frac{n}{2}\right).$$ There goes your desired result: $$I_n = \pi^{n/2}.$$