I learned that a complex integral can be computed by using Beta function. I can compute $\begin{align*} \int_{\infty}^{\infty}(1+x^{2})^{-n}\mathrm{d}x. \end{align*}$ by using the following relation. $$ \begin{align*} B\Big(\frac{1}{2}, n-\frac{1}{2}\Big) &= \int_{0}^{\infty} \frac{t^{-1/2}}{(1+t)^{n}}\mathrm{d}t \\ & = \int_{0}^{\infty} \frac{1/x}{(1+x^{2})^{n}}2x\mathrm{d}x \quad (\because \text{put} \ t =x^{2}, dt = 2x\mathrm{d}x)\\ & = 2\int_{0}^{\infty} \frac{1}{(1+x^{2})^{n}}\mathrm{d}x\\ &= \int_{-\infty}^{\infty} \frac{1}{(1+x^{2})^{n}}\mathrm{d}x. \end{align*} $$ In this example, $x$ is a scalar. I would like to extend this to $z \in \mathbb{R}^{N}$. Specifically, I would like to compute $$\int_{\infty}^{\infty}(1+z^{T}z)^{-n}\mathrm{d}z.$$ Can I compute this using the same procedure with $t=z^{T}z$ as follows?
$$ \begin{align*} B\Big(\frac{1}{2}, n-\frac{1}{2}\Big) &= \int_{0}^{\infty} \frac{t^{-1/2}}{(1+t)^{n}}\mathrm{d}t \\ & = \int_{0}^{\infty} \frac{z^{-1}}{(1+z^{T}z)^{n}}2z\mathrm{d}z \quad (\because \text{put} \ t =z^{T}z, dt = 2z\mathrm{d}z)\\ & = 2\int_{0}^{\infty} \frac{1}{(1+z^{T}z)^{n}}\mathrm{d}z\\ &= \int_{-\infty}^{\infty} \frac{1}{(1+z^{T}z)^{n}}\mathrm{d}z. \end{align*} $$
You are interested in computing $$\int_{\mathbb{R}^d} \frac{1}{(1+|x|^2)^n}\,dx = \omega_{d-1} \int_0^\infty \frac{r^{d-1}}{(1+r^2)^n}\,dr = \frac{1}{2}\omega_{d-1} \int_0^\infty \frac{r^{d/2-1}}{(1+r)^n} \,dr=\frac{\omega_{d-1}}{2}B\left(d/2, n-d/2\right),$$ with $\omega_{d-1}$ the area of $S^{d-1}$. Notice that the integral is well-defined only if $n>d/2$.
EDIT: changed $dx$ to $dr$.