In Pattern Recognition and Machine Learning by Bishop exercise 1.18, the author states the following relation:
$\prod_{i=1} \int_{-\infty}^{\infty} e^{-x_{i}^2} dx_{i}= S_{D} \int_{0}^{\infty} e^{-r^2} r^{D-1}dr$
where $S_{D}$ is the surface area of the unit sphere in D dimensions. How is this expression derived?
I know how to show that the integral on the left evaluates to $\pi$ for $n=2$ and therefore for $n=1$ it equals $\sqrt{\pi}$. Multiplying $\sqrt{\pi}$, $D$ times gives $\pi^{D/2}$.
On the right side of the expression in 2 dimensions I understand that we would have the double integral:
$\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}rdrd\theta$
However, I am unsure why one could rewrite this as:
$S_{D}\int_{0}^{\infty}e^{-r^2}rdr$
Does this only apply because we are dealing with radius = 1?
Additionally I am unsure how to generalize the expression to any number of dimensions:
$S_{D} \int_{0}^{\infty} e^{-r^2} r^{D-1}dr$