# 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 $$d=3-\epsilon$$): $$\int_{0}^{\infty}dp\frac{p^2}{p^2+m^2}$$ 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 $$\frac{2\pi^{d/2}}{\Gamma(d)}.$$ 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.

• I have seen conventions in dim. reg. using both $d=4-2\epsilon$ and $d=4-\epsilon$, but I rarely see one ever use $d=3-\epsilon$. I'm assuming you're computing a 2+1 dimensional integral rather than a 3+1? – Triatticus Mar 21 at 20:10
• @Triatticus, yes. The question is about DimReg in $2+1$ or in $2n+1$ – Artem Alexandrov Mar 21 at 20:58
• I see, then as you've answered that is the source of your issue, glad you sorted it out quickly. – Triatticus Mar 21 at 21:00

DimReg replaces any even divergnces (log, quadratic, etc.) by pole $$1/\epsilon$$. For any odd divergence, DimReg doesn't give the correct answer. This problem can be solved by PDS (Power Divergence Substruction) which is discussed in the following paper.