TITLE: Prove the following beta integral identity
QUESTION [10 upvotes]: I am looking at the following exercise from Grafakos' classical fourier analysis. For $0 < \alpha_1, \alpha_2 < n, \alpha_1 + \alpha_2 > n$, I want to prove that $$\int_{\mathbb{R}^n} \frac{1}{\left| x-t \right|^{\alpha_1} \left| y -t \right|^{\alpha_2}} dt = \pi^{\frac{n}{2}} \frac{\Gamma \left( \frac{n-\alpha_1}{2} \right) \Gamma \left( \frac{n- \alpha_2}{2} \right) \Gamma \left( \frac{\alpha_1 + \alpha_2 -n}{2} \right)}{\Gamma \left( \frac{\alpha_1}{2} \right) \Gamma \left( \frac{\alpha_2}{2} \right) \Gamma \left( n - \frac{\alpha_1 + \alpha_2}{2} \right)}\left| x -y \right|^{n- \alpha_1 - \alpha_2}.$$ 
I've been able to do most of the exercises in Grafakos, but don't even know where to start on this one.

REPLY [4 votes]: @Dap's answer provides a fundamental understanding on the identity. So let me safely ignore the issue of intuition and just supply a brutal force computation.
Before starting our computation, we remark some identities to be utilized:
1. Variant of the gamma function integral. For $\alpha > 0$ and $s > 0$ we have
$$ \frac{1}{s^{\alpha}} = \frac{2}{\Gamma(\frac{\alpha}{2})} \int_{0}^{\infty} \lambda^{\alpha-1}e^{-s^2\lambda^2} \, d\lambda. \tag{1}$$
2. Beta function integral. For $\alpha, \beta > 0$ we have
$$ \int_{0}^{\frac{\pi}{2}} \sin^{\alpha-1}\theta \cos^{\beta-1}\theta \, d\theta = \frac{\Gamma(\frac{\alpha}{2})\Gamma(\frac{\beta}{2})}{\Gamma(\frac{\alpha+\beta}{2})}. \tag{2}$$
Now let us begin our computation. Let $I$ denote the integral in question. Using $\text{(1)}$, we may write
$$ I = \frac{4}{\Gamma(\frac{\alpha_1}{2})\Gamma(\frac{\alpha_2}{2})} \int_{0}^{\infty} d\lambda_1 \int_{0}^{\infty} d\lambda_2 \int_{\mathbb{R}^n} \lambda_1^{\alpha_1-1}\lambda_2^{\alpha_2-1} e^{-\lambda_1^2 |x-t|^2 - \lambda_2^2|y-t|^2}. \tag{*} $$
(Interchanging the order of integration is justified since the integrand is positive; you may invoke Tonelli's theorem.) Now completing the square of the exponent
$$ -\lambda_1^2 |x-t|^2 - \lambda_2^2|y-t|^2
= -(\lambda_1^2 + \lambda_2^2)\left|t - \frac{\lambda_1^2 x+ \lambda_2^2 y}{\lambda_1^2+\lambda_2^2} \right|^2
- \frac{\lambda_1^2 \lambda_2^2}{\lambda_1^2+\lambda_2^2} |x - y|^2 $$
and applying the gaussian integral, $\text{(*)}$ simplifies to
$$ I = \frac{4\pi^{n/2}}{\Gamma(\frac{\alpha_1}{2})\Gamma(\frac{\alpha_2}{2})} \int_{0}^{\infty} d\lambda_1 \int_{0}^{\infty} d\lambda_2 \frac{\lambda_1^{\alpha_1-1}\lambda_2^{\alpha_2-1}}{(\lambda_1^2 + \lambda_2^2)^{n/2}} \exp\left\{ - \frac{\lambda_1^2 \lambda_2^2}{\lambda_1^2+\lambda_2^2} |x - y|^2\right\}. $$
Then applying the polar coordinates change $(\lambda_1, \lambda_2) = r(\cos\theta, \sin\theta)$, this further simplifies to
\begin{align*}
I
&= \frac{4\pi^{n/2}}{\Gamma(\frac{\alpha_1}{2})\Gamma(\frac{\alpha_2}{2})} \int_{0}^{\frac{\pi}{2}} d\theta \left( \int_{0}^{\infty} dr \, r^{\alpha_1+\alpha_2-n-1} e^{- r^2|x-y|^2 \sin^2\theta \cos^2\theta} \right)  \cos^{\alpha_1 -1}\theta \sin^{\alpha_2-1}\theta \\
&\stackrel{(1)}{=} \frac{2\pi^{n/2}}{\Gamma(\frac{\alpha_1}{2})\Gamma(\frac{\alpha_2}{2})} \int_{0}^{\frac{\pi}{2}} d\theta \left( \frac{\Gamma(\frac{\alpha_1+\alpha_2-n}{2})}{|x-y|^{\alpha_1+\alpha_2-n}\sin^{\alpha_1+\alpha_2-n}\theta \cos^{\alpha_1+\alpha_2-n}\theta} \right)  \cos^{\alpha_1 -1}\theta \sin^{\alpha_2-1}\theta\\
&\stackrel{(2)}{=} \pi^{n/2} \frac{\Gamma(\frac{\alpha_1+\alpha_2-n}{2}) \Gamma(\frac{n-\alpha_1}{2})\Gamma(\frac{n-\alpha_2}{2})}{\Gamma(\frac{\alpha_1}{2})\Gamma(\frac{\alpha_2}{2})\Gamma(n-\frac{\alpha_1+\alpha_2}{2})} |x-y|^{n-\alpha_1-\alpha_2}.
\end{align*}