The following integral was integrated using Mathematica:
$$\int_{-\infty}^{\infty} d\tau \frac{1}{(x^2+\tau^2)^a} = \frac{\Gamma(a-1/2)}{\Gamma(a)}\frac{\sqrt{\pi}}{|x|^{2 a -1}} \tag{1}$$
Mathematica says that this formula holds if $x^2\geq 0$ and $a>1/2$. I have the feeling that Mathematica gives the second condition to prevent divergent integrals ($\Gamma(\epsilon) \to \infty$ for $\epsilon \to 0$). However in my case, this integral showed up in the context of dimensional regularization, and $a=a(\omega)$ with $d=2\omega \to 4$ the number of dimensions. I am interested in values of $a\leq 1/2$, and would like to keep the divergence in the form of a gamma function as in $(1)$. Can I do that, or is the result of the integral something else for $a\leq 1/2$?