# Dimensional regularization and expansion of gamma function

In my calculations, I used dimensional regularization, i.e. replace $$d\rightarrow d-\epsilon$$ and calculated the divergent integral. Then, I would like to expand the answer into seriers by $$\epsilon$$ around $$\epsilon=0$$. But I obtained strange result. I start from the following integral (where I denote $$d=3-\epsilon$$): $$\int_{0}^{\infty}dp\frac{p^{3-1}}{p^2+m^2},$$ which is divergent. Then, I have calculated the integral $$I(\epsilon)=\int_{0}^{\infty}dp\frac{p^{d-1}}{p^2+m^2}=\frac{m^{d-2}}{2}\Gamma(d/2)\Gamma(1-d/2),$$ which is convergent for $$d<2$$. Also, I also have integral over angles, which is in $$d$$-dimensional case can be written as $$\frac{2\pi^{d/2}}{\Gamma(d/2)}.$$ So, my final answer is $$I(\epsilon)\propto\Gamma(\epsilon/2-1/2).$$ Using Wolfram Mathematia, I find the expansion around $$\epsilon=0$$. My expectation was that the divergence of my integral will be appear like a pole, $$1/\epsilon$$. But from the expansion I see no one singular term.

The key problem is that dimensional regularization is insensitive to higher divergences, i.e. if the integral diverges linear (my case) the divergence is not a pole $$1/\epsilon$$.
The question is near to close. Now the main problem is convergence of considered integral. Indeed, one can calculate the integral in $$d$$-dimensional space and expressed the answer in term of Euler $$\beta$$-function. Then, if one set $$d=3$$ the answer is finite while the initial integral is linearly divergent.