Integral of $\exp(-\|x\|_p)$

Question

$\newcommand{\RR}{\mathbb{R}}$ Let $1 \leq p < \infty$ and let $d$ be a positive integer. I want to show that $$I := \int_{\RR^d} \exp\left(-\|x\|_p\right) dx = \frac{(2\Gamma(1/p))^d}{p^{d-1}}\frac{\Gamma(d)}{\Gamma(d/p)}.$$ Due to the $L^p$ norm, the integrand is not radially symmetric. If I convert to spherical coordinates I end up with $$I = \Gamma(d) \int_{S^{d-1}} \frac{d\sigma(x)}{\|x\|_p^d} $$ where $\sigma$ is the spherical measure. How should I proceed from here? Or alternatively how should I compute this integral? I would appreciate detailed calculations!

Answer 1

A simple approach is to see that, for any $f$ such that the integral on the LHS below exists, we have $$\int_{\mathbb{R}^d}f(\|x\|_p)\,dx=K_{p,d}\int_0^\infty t^{d-1}f(t)\,dt,\tag{*}\label{maineq}$$ where $K_{p,d}$ doesn't depend on $f$ (say, we substitute $x=tg(y)$ where $x=g(y)$ is a parameterization of $\|x\|_p=1$; the Jacobian of such a substitution is of the form $t^{d-1}h(y)$ with some $h$ not depending on $t$).

To find $K_{p,d}$, we choose $f$ so that both sides of \eqref{maineq} are easy to compute. Take $f(t)=e^{-t^p}$: $$\int_{\mathbb{R}^d}f(\|x\|_p)\,dx=\left(\int_\mathbb{R} e^{-|x|^p}\,dx\right)^d=\big(2\Gamma(1/p)/p\big)^d,\\\int_0^\infty t^{d-1}f(t)\,dt=\int_0^\infty t^{d-1}e^{-t^p}\,dt=\Gamma(d/p)/p,$$ so that $K_{p,d}=\frac{2^d\Gamma^d(1/p)}{p^{d-1}\Gamma(d/p)}$. To get the answer, it remains to put $f(t)=e^{-t}$ in \eqref{maineq}.