A multidimensional quantum-type integral I can compute the following integral in a few lines via Schwinger parametrization, but the form of the integral is so simple, and so reminiscent of the result that multiplying a 1D function by powers of $x$ modifies the Fourier transform by differentiation/integration, that I feel there's a neat way of doing it that I've missed, having thought about it for a little bit. Am I right?
$$\int \mathrm d^3 \mathbf p' \frac 1 {(\mathbf p - \mathbf p')^2} \frac 1 {(\mathbf p'^2+\alpha^2)^2} = \frac {\pi^2} {\alpha(\mathbf p^2+\alpha^2)}$$
To be clear, what I want is a simple argument that tells me why the form is so simple (including a reason for the recurrence of the $\mathbf p^2+\alpha^2$ term), or a decent reason to believe this is just a fluke.
 A: Not sure this helps, but my first reaction to this is to integrate in spherical coordinates:
$$\int_{\mathbb{R}} \mathrm d^3 \mathbf p' \frac 1 {(\mathbf p - \mathbf p')^2} \frac 1 {(\mathbf p'^2+\alpha^2)^2} = 2 \pi \int_0^{\infty} dp \, \frac{p'^2}{(p'^2+\alpha^2)^2} \, \int_0^{\pi} d\theta \frac{\sin{\theta}}{p'^2+p^2-2 p p' \cos{\theta}}$$
The integral over $\theta$ is relatively simple through e.g., subbing $u=\cos{\theta}$:
$$\int_0^{\pi} d\theta \frac{\sin{\theta}}{p'^2+p^2-2 p p' \cos{\theta}} = \frac{1}{2 p p'} \log{\left [\left (\frac{p'+p}{p'-p} \right )^2\right ]}$$
Then the integral is now a single integral equal to
$$\frac{\pi}{p} \int_0^{\infty} dp \, \frac{p'}{(p'^2+\alpha^2)^2} \log{\left [\left (\frac{p'+p}{p'-p} \right )^2\right ]}$$
The integrand has a logarithmic singularity at $p'=p$; while this is integrable, it will cause a little grief upon integration by parts, as it will reveal an integral that inherently does not converge:
$$2 \pi \int_0^{\infty} dp \, \frac{1}{p'^2+\alpha^2} \frac{1}{p^2-p'^2}$$
I will continue along, however, assuming that we are interested in the Cauchy principal value.  Setting $p'=\alpha \tan{\phi}$, the integral becomes
$$\frac{2 \pi}{\alpha} \int_0^{\pi/2} \frac{d\phi}{p^2-\alpha^2 \tan^2{\phi}} $$
Use symmetry to express this integral in terms of a contour integral in the complex plane upon subbing $z=e^{i \phi}$:
$$\frac{-i \pi}{2 \alpha} \oint_{|z|=1} \frac{dz}{z} \frac{(z^2+1)^2}{(p^2+\alpha^2) z^4+2 (p^2-\alpha^2) z^2 + (p^2+\alpha^2)}$$
The poles of the integrand are at $z=0$ and $z=\pm(p\pm i \alpha)/\sqrt{p^2+\alpha^2}$.  The latter poles are on the unit circle, as expected, but because we are interested in the Cauchy PV, we may deform the circle so as to avoid these poles; the contributions from the deformations do not contribute to the contour integral.  Thus we are left with the pole at $z=0$; the contour integral is $i 2 \pi$ times the residue of the integrand at $z=0$, or
$$i 2 \pi \frac{-i \pi}{2 \alpha} \frac{1}{p^2+\alpha^2} = \frac{\pi^2}{\alpha (p^2+\alpha^2)}$$
as was asserted.
