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.

  • $\begingroup$ 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? $\endgroup$
    – Triatticus
    Mar 21 '19 at 20:10
  • $\begingroup$ @Triatticus, yes. The question is about DimReg in $2+1$ or in $2n+1$ $\endgroup$ Mar 21 '19 at 20:58
  • $\begingroup$ I see, then as you've answered that is the source of your issue, glad you sorted it out quickly. $\endgroup$
    – Triatticus
    Mar 21 '19 at 21:00

I would like to present the answer.

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.


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