Prove: $\int_{\mathbb{R}^n}e^{-|\mathbf{x}|^n}\,d^n\mathbf{x} = \operatorname{vol}(B(0,1))$

Question

Prove:

$$\int_{\mathbb{R}^n}e^{-|\mathbf{x}|^n}\,d^n\mathbf{x} = \operatorname{vol}(B(0,1))$$

where $B(0,1)$ is the unit ball in $\mathbb{R}^n$. I'm not sure how to approach this and would appreciate any help.

Answer 1

This can be done using spherical coordinates in $n$ dimensions.

$$\begin{align} x_1 &= r\cos(\phi_1) \\ x_2 &= r\sin(\phi_1)\cos(\phi_2)\\ x_3 &= r\sin(\phi_1)\sin(\phi_2)\cos(\phi_3)\\ &\ \ \vdots\\ x_{n-1} &= r\sin(\phi_1)\sin(\phi_2)\dots\sin(\phi_{n-2})\cos(\phi_{n-1})\\ x_{n} &= r\sin(\phi_1)\sin(\phi_2)\dots\sin(\phi_{n-2})\sin(\phi_{n-1})\\ \end{align} $$

The Jacobian is $$d^n \mathbf x = r^{n-1} \sin^{n-2}(\phi_{1})\sin^{n-3}(\phi_{2})\dots\sin(\phi_{n-2})\ dr\ d\phi_1\dots d\phi_{n-1}$$

$$\begin{align} I &= \int_{\mathbb R^n} e^{-|\mathbf x|^n}\ d^n|\mathbf x| \\ &= \int_{\substack{r\in [0,\infty)\\ \phi_1,\dots\phi_{n-2}\in [0,\pi]\\ \phi_{n-2}\in[0,2\pi]}}r^{n-1}e^{-r^n}\sin^{n-2}(\phi_{1})\sin^{n-3}(\phi_{2})\dots\sin(\phi_{n-2})\ dr\ d\phi_1\dots d\phi_{n-1} \end{align} $$

Let $s = 1-e^{-r^n}$, then $ds = nr^{n-1}e^{-r^n}\ dr$

$$\begin{align} I &= \int_{\substack{s\in [0,1]\\ \phi_1,\dots\phi_{n-2}\in [0,\pi]\\ \phi_{n-2}\in[0,2\pi]}} \frac{1}{n}\sin^{n-2}(\phi_{1})\sin^{n-3}(\phi_{2})\dots\sin(\phi_{n-2})\ ds\ d\phi_1\dots d\phi_{n-1} \\ &= \int_{\substack{t\in [0,1]\\ \phi_1,\dots\phi_{n-2}\in [0,\pi]\\ \phi_{n-2}\in[0,2\pi]}} t^{n-1}\sin^{n-2}(\phi_{1})\sin^{n-3}(\phi_{2})\dots\sin(\phi_{n-2})\ ds\ d\phi_1\dots d\phi_{n-1}\\ &= \int_{B(0,1)}\ d^n\mathbf x_n\\ I &= V(B(0,1)) \end{align} $$

Where $t$ is our new "radius" s.t $t^n = s$ and $nt^{n-1}\ dt=ds$