Consider the following integral:

\begin{equation} \int \frac{d^3 k}{(2\pi)^3} \frac{1}{k^3} \end{equation}

In dim-reg, such integrals evaluate to $0$. However, if we instead consider

\begin{eqnarray} \int \frac{d^3 k}{(2\pi)^3} \frac{(p+k)^2}{k^3(p+k)^2} &=& \int dx \int \frac{d^3 k}{(2\pi)^3} \frac{k^2 + (1-x)^2 p^2}{(k^2+x(1-x)p^2)^{5/2}} \frac{\Gamma(\frac{5}{2})}{\Gamma(\frac{3}{2})}(1-x)^{1/2} \\ &=& \int dx \frac{\mu^{2\epsilon}}{(4\pi)^{3/2 - \epsilon}} \left[\frac{\Gamma(\epsilon)\Gamma(\frac{5}{2})}{\Gamma(\frac {3}{2})\Gamma(\frac{5}{2})} \left(\frac{1}{x(1-x)p^2}\right)^\epsilon + \text{finite}\right] \frac{3}{2} (1-x)^{1/2}\\ &=& \frac{1}{4\pi^2} \Gamma{(\epsilon)} \mu^{2\epsilon} p^{-2\epsilon} + \text{const} \end{eqnarray} where $d = 3-2\epsilon$.

Even though the integrals are the same (after canceling out $(p+k)^2$), they give different results when evaluated under dim-reg. My understanding of this is that when one did the Feynman parameterization, an IR regulator was secretly introduced when interchanging the $dx$ and the $d^3 k$.

  1. How exactly is the IR being regulated in the second case?
  2. The fact that the first case evaluates to $0$ is due to a cancellation between UV and IR divergences. Is there an easy way to evaluate only the UV divergence of the first integral within dim-reg?
  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/136424 $\endgroup$
    – innisfree
    Commented Mar 29, 2017 at 4:54
  • $\begingroup$ Hi, it's been a long time since I've done dim-regularized integrals. Can you please explain to me why there is what there is in the numerator after Feynman parametrization? $\endgroup$ Commented Mar 30, 2017 at 2:08
  • $\begingroup$ @SolenodonParadoxus The Feynman parametrization formula is $\frac{1}{A^\alpha B^\beta} = \int dx \frac{1}{(xA + (1-x)B)^{\alpha + \beta}} x^{\alpha -1} (1-x)^{\beta - 1} \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}$. There is a shift $k \rightarrow k-xp$, and the $k\cdot p$ term evaluates to $0$ so it is dropped. $\endgroup$
    – Aaron
    Commented Mar 30, 2017 at 2:14
  • $\begingroup$ @Aaron: I think the denominator should be to the power of $5/2$. I get a different numerator structure too. $\endgroup$
    – CAF
    Commented Mar 30, 2017 at 11:45
  • 2
    $\begingroup$ Having a semi-detailed answer would be nice. I'll renew the bounty if it expires without an answer. $\endgroup$
    – Helen
    Commented Apr 10, 2017 at 11:07

1 Answer 1


You have to be careful when performing certain operations in dimensional regularization, and you always have to check the convergence.

What meaning does an integral like $$ \int\frac{d^Dk}{(2\pi)^D}\frac{1}{k^2+m^2} $$ have in dimensional regularization? Here I take $m>0$, strictly. If you take $D$ as a natural number, the above integral is just a regular integral, that you can write by performing the angular integral first: $$ \int\frac{d^Dk}{(2\pi)^D}\frac{1}{k^2+m^2}=\frac1{2^{D-1}\pi^{\frac D2}\Gamma\left(\frac D2\right)}\int_0^\infty dk\frac{k^{D-1}}{k^2+m^2}. $$ Now, what is done in dimensional regularization is to aknowledge the fact that the RHS makes sense for $D$ complex, barring divergence, so one "analitically continues" the LHS to complex values of $D$. But complex analysis tells us that, in order to have an analytic continuation, the function (of $D$) that we wish to continue must be well defined and not divergent in an open set of the complex plane.

This happens in this case. The integrand goes to $k^{D-1}$ when $k$ is near zero (the presence of a non null $m^2$ is crucial for this fact), so you have convergence around zero when $\text{Re}(D)-1>-1$, while near $\infty$ you have that the integrand goess to $k^{D-3}$, so you have convergence when $\text{Re}(D)-3<-1$. In total, the integral is convergent in the strip $0<\text{Re}(D)<2$, and analytic in that strip. So, you can perform analytical continuation in the complex plane and obtain a formal expression for each complex value of $D$, to use in calculations.

If, however, $m=0$ you can run in some problems. The integral $$ \int\frac{d^Dk}{(2\pi)^D}\frac{1}{k^3}=\frac{1}{2^{D-1}\pi^{\frac D2}\Gamma\left(\frac D2\right)}\int_0^\infty k^{D-4}dk $$ does not converge, as for convergence in zero you need $\text{Re}(D)-4>-1$ and for convergence at infinity you need $\text{Re}(D)-4<-1$, so no value of $D$ can satisfy both conditions. As the object is still undefined, we could define it by manipulating it: $$ \int\frac{d^Dk}{(2\pi)^D}\frac{1}{k^3}\frac{k^2+\delta^2}{k^2+\delta^2}:=\int\frac{d^Dk}{(2\pi)^D}\frac{1}{k(k^2+\delta^2)}+\delta^2\int\frac{d^Dk}{(2\pi)^D}\frac{1}{k^3(k^2+\delta^2)}. $$ Note the definition sign $:=$, I'm not using linearity of the integral (as the integral is not defined) but I'm just defining that integral as sum of two integrals. The LHS still is the integral of $\frac 1{k^3}$, and the successive fraction is there just for convenience, the RHS is a sum of two integrals that have two (disjoint) strips of convergence. But both integrals have strips of convergence, so they can be both analitically continued. Both terms of the sum are well defined, and the sum is well defined, so you can assign a value to the integral of $\frac 1{k^3}$. We wish that value to be $\delta$ independent, obviously: $\delta$ is $\textit{not}$ a regulator here. As those integrals are convergent somewhere (in an open set), you can do all sort of manipulations, like exchanging order of integration when using the Feynmann trick.

Now, let's compute this sum by using the Feynmann trick, to see if we get zero. For the first integral, you have $$ \frac{\Gamma\left(\frac32\right)}{\Gamma\left(\frac12\right)}\int_0^1 dx\int\frac{d^Dk}{(2\pi)^D}\frac{1}{\sqrt{x}(k^2+\delta^2(1-x))^{\frac32}}=\frac{\sqrt{\pi}}2\frac{\Gamma\left(\frac{3-D}{2}\right)}{(4\pi)^{\frac D2}\Gamma\left(\frac{3}{2}\right)}\delta^{D-3}\frac{ \Gamma \left(\frac{D-1}{2}\right)}{\Gamma \left(\frac{D}{2}\right)}. $$ Note that I've exchanged the order of integration, as there is a convergence strip of the integral. If you have doubts about this result, consult Appendix A of the reference at the end of the answer for useful formulas. Note that you have a definite value of the result when $1<\text{Re}(D)<3$, as the Euler gamma is always defined for arguments with positive real part. This is a confirmation of the fact that we are on the right track.

I will omit calculations for the second. At the end, you just have $$ \frac{3\sqrt\pi}{4}\frac{\Gamma\left(\frac{5-D}{2}\right)}{(4\pi)^\frac D2\Gamma\left(\frac 52\right)}\delta^{D-3}\frac{\Gamma\left(\frac{D-3}{2}\right)}{\Gamma\left(\frac D2\right)}. $$

Now we sum. Note that here you have to use the properties of $\Gamma$. The sum is $$ \frac{\sqrt\pi}{2}\frac{1}{(4\pi)^{\frac D2}\Gamma\left(\frac D2\right)}\left(\Gamma\left(\frac{5-D}{2}\right)\Gamma\left(\frac{D-3}{2}\right)+\Gamma\left(\frac{3-D}{2}\right)\Gamma\left(\frac{D-1}{2}\right)\right)\delta^{D-3}. $$ You can simply prove through the properties of $\Gamma$ that the term in parentheses is $0$. So, the integral of $\frac 1{k^3}$ in dimensional regularization is $0$, in this sense. Notice that it is exaclty $0$, no limit and no $\delta$ dependence.

To summarize, your error lies in the exchange of the $dx$ and $dk$ integrals. To do this exchange, you have to find an open set in the complex plane where the integral makes sense and is convergent. In your case, you could try by expanding the numerator and "splitting" (improper terminology, as you are defining, not splitting) the integral: each term of the sum has a strip of convergence, so you can use the Feynmann trick. Try to do that and see if you can obtain zero.

My reference for this answer is the book by Damiano Anselmi, "Renormalization", freely downloadable from his site, http://renormalization.com/ (under Documents->Books. Yeah, he owns that domain!). Chapter 2's beginning introduces dimensional regularization.

  • $\begingroup$ I don't have the knowledge to say if the answer is correct in its details (I'll hopefully have some more after studying it and the reference). But it obviously deserves a bounty. $\endgroup$
    – Helen
    Commented Apr 19, 2017 at 16:56

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