How do you do multivariate integration by substitution with different dimensions?
I am reading a paper doing this change of variable with $x,t \in \mathbb{R}^n$:
$$ \frac{1}{(2 \pi)^{n / 2}} \int_{\mathbb{R}^n}\|(x-t)\|_2 \exp \left(-\frac{1}{2}\|x-t\|_2^2\right) d t \\ $$ to $$ \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty}|s| \exp \left(-\frac{1}{2} s^2\right) d s $$
To my understanding, $s(x,t)=\|x-t\|_2^2$, $s:\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$.
Two sources of confusion:
- For integration by substitution, I need $|det(J)|=|det(\nabla_t^Ts)|$, but $\nabla_t^Ts$ is not square and thus the determinant doesnt exist.
- Wiki says that $s$ needs to be injective, which it isn't.
Question:
How do I get from the first to the second line?
I have to say, I’m not sure how they arrived at the final integral because the initial integral, call it $I_n$, depends on $n$ non-trivially (as shown below) whereas the final integral doesn’t.
First make a translation $\Bbb{R}^n\to\Bbb{R}^n$, $t\mapsto z=t+x$, and then change to polar coordinates to get \begin{align} I_n&= \frac{1}{(2\pi)^{n/2}}\int_{\Bbb{R}^n}\|z\|\exp\left(-\frac{\|z\|^2}{2}\right)\,dz\\ &=\frac{1}{(2\pi)^{n/2}}\int_0^{\infty}\int_{S^{n-1}}s\exp\left(-\frac{s^2}{2}\right)\cdot s^{n-1}\,d\Omega\,ds\\ &=\frac{A_{n-1}}{(2\pi)^{n/2}}\int_0^{\infty}s^ne^{-s^2/2}\,ds, \end{align} where $A_{n-1}=\int_{S^{n-1}}1\,d\Omega=\frac{2\pi^{n/2}}{\Gamma(n/2)}$ is the surface ‘area’ of the unit sphere $S^{n-1}\subset\Bbb{R}^n$ (which is equal to $2\pi$ if $n=2$, and $4\pi$ if $n=3$ etc). Now, we make a change of variables $s=\sqrt{2y}$, $ds=\frac{dy}{\sqrt{2y}}$ to get \begin{align} I_n&=\frac{A_{n-1}}{(2\pi)^{n/2}}\int_0^{\infty}\left(\sqrt{2y}\right)^{n-1}e^{-y}\,dy\\ &=\frac{A_{n-1}}{\sqrt{2}\pi^{n/2}}\int_0^{\infty}y^{\frac{n-1}{2}}e^{-y}\,dy\\ &=:\frac{A_{n-1}}{\sqrt{2}\pi^{n/2}}\Gamma\left(\frac{n-1}{2}+1\right)\\ &=\frac{\frac{2\pi^{n/2}}{\Gamma(n/2)}}{\sqrt{2}\pi^{n/2}}\Gamma\left(\frac{n+1}{2}\right)\\ &=\sqrt{2}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}, \end{align} where we have used the definition of the Gamma function and the above exact value for the surface area $A_{n-1}$. Now, using some knowledge of the particular values of the Gamma function at positive half-integers, we can simplify this final expression to $I_n=\frac{(n-1)!!}{(n-2)!!}$.