I would like to compute the an integral of a function that takes in a vector, but got stuck. Would anyone tell me how to compute it?
Let $\boldsymbol{x}$ be a vector as below.
$$ \boldsymbol{x} = [x_{1}, \cdots, x_{n}]^{T} $$
Integral $I$ is given by the following equation with $i,j=1 \cdots n$, and $p\in\mathbb{R}$ $$I=\int_{\mathbb R^n}(1+||x||^{2})^{-p}x_{i}x_{j}\mathrm{d}\boldsymbol{x}$$
Let us consider the more general case $$T_{ij}:=\int\limits_{\mathbb{R}^n}\! d^nx \, x_i x_j \, f(\Vert x \Vert), $$ with an arbitrary function $f$ such that the integrals exist. For any ${\rm SO}(n)$ matrix $R$ ($R^TR=R R^T=\mathbf{1}_n, \, \det R =1$), the relation $$\sum\limits_{k=1}^n\sum\limits_{\ell=1}^n R_{ik} R_{j\ell} T_{k\ell}=\int\limits_{\mathbb{R}^n}\! d^n x \, (R x)_i(Rx)_j \, f(\Vert x \Vert) = \int\limits_{\mathbb{R}^n} \! d^n y \, y_i y_j \, f(\Vert y\Vert)=T_{ij}$$ holds, where the variable transformation $y=Rx$ together with $\det R =1$ and $\Vert Rx \Vert =\Vert x \Vert$ was performed. Written in matrix notation, we have found $$ R T R^T=R \quad \Leftrightarrow \quad R T =TR,$$ i.e. $T$ commutes with all elements of ${\rm SO}(n)$ and Schur tells us that $T$ must be proportional to the unit matrix, $ T= \lambda \mathbf{1}_n$. Taking the trace, we find $$\lambda = \frac{1}{n}\int\limits_{\mathbb{R}^n}\! d^nx \, \Vert x \Vert ^2 f(\Vert x \Vert) =\frac{V_{{\rm S}^{n-1}}}{n} \int\limits_0^\infty\! dr \, r^{n-1} r^2 f(r), $$ where $V_{{\rm S}^{n-1}}=2 \pi^{n/2}/\Gamma(n/2)$ is the volume of the $(n-1)$-dimensional sphere ${\rm S}^{n-1}$, giving the final result $$ T_{ij}= \frac{ 2 \pi^{n/2}\delta_{ij}}{n \, \Gamma(n/2)} \int\limits_0^\infty \! dr \, r^{n+1} f(r).$$ The further steps for the special case $f(r)=(1+r^2)^{-p}$ (with $p> 1+n/2$) are left as an exercise.