What's the intuition behind change of variables?

Question

I'm reading a derivation that confidently and without further comment asserts $$ \int_{x_1=-\infty}^\infty \cdots \int_{x_n=-\infty}^\infty e^{-\sum_i x_i^2}\,dx_n\cdots dx_1 = \int_{r=0}^\infty e^{-r^2}\,\frac{d\text{Volume}(S_r^n)}{dr}\,dr\,. $$ (Where $S_r^n$ is the $n$-dimensional hypersphere with radius $r$.)

Why is this obvious? The only way I know how to do a change of variables is by finding an explicit transformation whose Jacobian I can then derive and stick into the integral together with the new variables. And I'm not even sure how exactly polar coordinates should generalize to $n$ dimensions. Clearly, the author of this derivation was able to sidestep all that with some geometric intuition that I'm not sharing.

Answer 1

Consider the two-dimensional case, using polar coordinates: $$ I = \int_{x=-\infty}^{x=\infty} \int_{y=-\infty}^{y=\infty} e^{-(x^2+y^2)}\,dy\,dx = \int_{r=0}^{r=\infty} \int_{\theta=0}^{\theta=2\pi} e^{-r^2}r\,d\theta\,dr. $$ We can take a constant factor out of the inner integral: $$ I = \int_{r=0}^{r=\infty} \left(e^{-r^2}\int_{\theta=0}^{\theta=2\pi} r\,d\theta\right)\,dr. $$ Now, $r\,d\theta$ is just an element of arc length, that is, $\int_{\theta=0}^{\theta=2\pi} r\,d\theta$ measures the circumference of a circle of radius $r,$ which is the boundary of a disk of radius $r,$ which has the same measure as the rate of change of the area of the disk with respect to $r.$ Putting this in the notation used in the question, $$ \int_{\theta=0}^{\theta=2\pi} r\,d\theta = \frac{d\text{Volume}(S_r^2)}{dr} = 2\pi r. $$ In the three-dimensional case, the inner integral would be $$ \int_{\phi=0}^{\phi=\pi} \int_{\theta=0}^{\theta=2\pi} r^2 \sin\phi\,d\theta\,d\phi = \frac{d\text{Volume}(S_r^3)}{dr} = 4\pi r^2. $$ We can explicitly do higher-dimensional integrals the same way, but the effort we have to make in order to correctly represent the volume element of the inner integral in $n$ dimensions merely obscures the fact that in every case the inner integral is just integrating $1$ over the boundary of an $n$-dimensional ball, and the integral is just the measure of the boundary, which is $\frac{\text{Volume}(S_r^n)}{dr}.$

A possible source of confusion is that this author uses the notation $\text{Volume}(S_r^n)$ where the volume measure of an $n$-dimensional ball is required, but calls $S_r^n$ an "$n$-dimensional hypersphere," which seems inconsistent with the way other authors use that term. We have to resolve that ambiguity in the intended way to see what the factor $\frac{\text{Volume}(S_r^n)}{dr}$ means in this context.

Answer 2

Change of variables works in a quite general way. If we write $B_r = \left\{ x \in \mathbb{R}^n : |x| \leq r \right\}$, then

\begin{align*} \int_{\mathbb{R}^n} e^{-|x|^2} \, \operatorname{Vol}(dx) &= \sum_{i=1}^{\infty} \int\limits_{\frac{i-1}{N} < |x| \leq \frac{i}{N}} e^{-|x|^2} \, dx\\ &\approx \sum_{i=1}^{\infty} e^{-(i/N)^2} \operatorname{Vol}\left(\left\{ x \in \mathbb{R}^n : \tfrac{i-1}{N} < |x| \leq \tfrac{i}{N}\right\}\right) \\ &\approx \int_{0}^{\infty} e^{-r^2} \, d\operatorname{Vol}\left(B_r\right) \\ &= \int_{0}^{\infty} e^{-r^2}\frac{d\operatorname{Vol}\left(B_r\right)}{dr} \, dr, \end{align*}

and this becomes exact as $N\to\infty$. You may think of this as a generalization of the shell method.