Gaussian Integral 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 $I^{3}, I^{4}, \dots, I^{n}$?
 A: 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 $|x| = \sqrt{x_1^2 + \cdots + x_n^2}$.
By spherical coordinate integration transform in $\mathbb{R}^n$:
$$
I_n  = \int_0^\infty\left\{\int_{\partial B(0,r)} e^{-|x|^2}\,dS\right\}\,dr.
$$
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:
$$
I_n = \pi^{n/2}.
$$
A: For three dimensions (if I figure out higher dimensions I will edit), one could try a similar argument by using spherical coordinates. If you can perform the integral, $$\int_0^\infty r^2 e^{-r^2} dr$$, then we can compute the desired integral. But This question addresses this. For this you need the spherical change of co-ordinates see this. You simply mimic the proof for the two dimensional case. I will leave the work to you.
Edit It seems in higher dimensions if one knows how to compute the integral, 
$$\int_0^\infty r^{n-1} e^{-r^2} dr$$ then we can compute the integral in essentially the same way. To see this you need higher dimensional spherical co-ordinates. For this see here.
