tun1ju2k1ki

2022-09-06

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

ordonansexa

Beginner2022-09-07Added 7 answers

As you have noticed, the integral can be transformed into

$${\int}_{0}^{\mathrm{\infty}}{d}^{n}y\frac{1}{(1+{y}^{T}y{)}^{\alpha}}.$$

Going to the spherical coordinates leads to

$$\frac{1}{{2}^{n}}\int d{\mathrm{\Omega}}_{n}{\int}_{0}^{\mathrm{\infty}}dr\frac{{r}^{n-1}}{(1+{r}^{2}{)}^{\alpha}},$$

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 $x=(1+{r}^{2}{)}^{-1}$, which leads to

$${\int}_{0}^{\mathrm{\infty}}dr\frac{{r}^{n-1}}{(1+{r}^{2}{)}^{\alpha}}=\frac{1}{2}{\int}_{0}^{1}{x}^{\alpha -n/2-1}(1-x{)}^{n/2-1}dx=\frac{\mathrm{\Gamma}(\alpha -n/2)\mathrm{\Gamma}(n/2)}{2\mathrm{\Gamma}(\alpha )},$$

where the expression for the Beta function was used in the last step. Putting the pieces together, you arrive at

$${\int}_{0}^{\mathrm{\infty}}{d}^{n}y\frac{1}{(1+{y}^{T}y{)}^{\alpha}}={\left(\frac{\pi}{4}\right)}^{n/2}\frac{\mathrm{\Gamma}(\alpha -n/2)}{\mathrm{\Gamma}(\alpha )}.$$

$${\int}_{0}^{\mathrm{\infty}}{d}^{n}y\frac{1}{(1+{y}^{T}y{)}^{\alpha}}.$$

Going to the spherical coordinates leads to

$$\frac{1}{{2}^{n}}\int d{\mathrm{\Omega}}_{n}{\int}_{0}^{\mathrm{\infty}}dr\frac{{r}^{n-1}}{(1+{r}^{2}{)}^{\alpha}},$$

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 $x=(1+{r}^{2}{)}^{-1}$, which leads to

$${\int}_{0}^{\mathrm{\infty}}dr\frac{{r}^{n-1}}{(1+{r}^{2}{)}^{\alpha}}=\frac{1}{2}{\int}_{0}^{1}{x}^{\alpha -n/2-1}(1-x{)}^{n/2-1}dx=\frac{\mathrm{\Gamma}(\alpha -n/2)\mathrm{\Gamma}(n/2)}{2\mathrm{\Gamma}(\alpha )},$$

where the expression for the Beta function was used in the last step. Putting the pieces together, you arrive at

$${\int}_{0}^{\mathrm{\infty}}{d}^{n}y\frac{1}{(1+{y}^{T}y{)}^{\alpha}}={\left(\frac{\pi}{4}\right)}^{n/2}\frac{\mathrm{\Gamma}(\alpha -n/2)}{\mathrm{\Gamma}(\alpha )}.$$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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