tun1ju2k1ki

2022-09-06

Solving ${\int }_{0}^{\mathrm{\infty }}\left(1+\frac{{y}_{1}^{2}+{y}_{2}^{2}+\cdots +{y}_{n}^{2}\right)}{\nu }\right)\mathrm{d}{y}_{1}\mathrm{d}{y}_{2}\cdots \mathrm{d}{y}_{n}$

ordonansexa

As you have noticed, the integral can be transformed into
${\int }_{0}^{\mathrm{\infty }}{d}^{n}y\frac{1}{\left(1+{y}^{T}y{\right)}^{\alpha }}.$
Going to the spherical coordinates leads to
$\frac{1}{{2}^{n}}\int d{\mathrm{\Omega }}_{n}{\int }_{0}^{\mathrm{\infty }}dr\frac{{r}^{n-1}}{\left(1+{r}^{2}{\right)}^{\alpha }},$
where the factor of $1/{2}^{n}$ is due to integrating over this fraction of the whole space and $\int d{\mathrm{\Omega }}_{n}$ is the volume of ${S}^{n-1}$, which is equal to
$\int d{\mathrm{\Omega }}_{n}=\frac{2{\pi }^{n/2}}{\mathrm{\Gamma }\left(\frac{n}{2}\right)},$
as can be shown for example by evaluating the integral $\int {d}^{n}y{e}^{-{y}^{T}y}$ both in Cartesian and spherical coordinates and comparing the two results.
To evaluate the simple integral over r, make the change of variables $x=\left(1+{r}^{2}{\right)}^{-1}$, which leads to
${\int }_{0}^{\mathrm{\infty }}dr\frac{{r}^{n-1}}{\left(1+{r}^{2}{\right)}^{\alpha }}=\frac{1}{2}{\int }_{0}^{1}{x}^{\alpha -n/2-1}\left(1-x{\right)}^{n/2-1}dx=\frac{\mathrm{\Gamma }\left(\alpha -n/2\right)\mathrm{\Gamma }\left(n/2\right)}{2\mathrm{\Gamma }\left(\alpha \right)},$
where the expression for the Beta function was used in the last step. Putting the pieces together, you arrive at
${\int }_{0}^{\mathrm{\infty }}{d}^{n}y\frac{1}{\left(1+{y}^{T}y{\right)}^{\alpha }}={\left(\frac{\pi }{4}\right)}^{n/2}\frac{\mathrm{\Gamma }\left(\alpha -n/2\right)}{\mathrm{\Gamma }\left(\alpha \right)}.$

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