I am using dimensional regularization to extract the divergence of some complicated integral. I work in $d=2\omega$ dimensions, with $\omega\approx 2$. After I extract the divergence, I have an expression of the form

$$f(\omega)\Gamma(\omega-2)\int_{-\infty}^{\infty}d\tau_3 d\tau_4 \frac{1}{(x_{13}^2)^{\omega-1}}\frac{1}{(x_{24}^2)^{\omega-1}}\frac{1}{(x_{34}^2)^{\omega-2}}\tag{1}$$

with $x_{ij}:=x_i-x_j$. Now I know how to compute

$$\int_{-\infty}^{\infty}d\tau_3 \frac{1}{(x_{13}^2)^{\omega-1}}\tag{2}$$

but the last factor spoils it. However, the integral seems finite, so if I send $\omega\to 2$ now, the last factor is simply $1$ and the integral is easy to compute.

Am I allowed to send $\omega\to 2$ for just one part of the integral, if the latter is finite? More generally, can I send $\omega\to 2$ for parts of a computation if they are finite in this number of dimensions?

Note that although this is a mathematical question, I felt that this was belonging to the physics page since (1) dimensional regularization is a tool that is used a lot in QFT, (2) the computation is directly related to a physics research, and other people probably thought about this question before in the physics community.

Clarification about the notation:

I forgot a few details about the remaining integrals: $x_{3\mu}$ and $x_{4\mu}$ are, respectively, defined as $(0,0,0,\tau_3)$ and $(0,0,0,\tau_4)$, while $x_1=(1,0,0,0)$ and $x_2=(x_2^1,x_2^2,0,0)$. Note that I work in Euclidean space. Thus, the integrals can be written as:

$$\int_{-\infty}^{\infty} d\tau_3 d\tau_4 \frac{1}{(x_1^2+\tau_3^2)^{\omega-1}}\frac{1}{(x_2^2+\tau_4^2)^{\omega-1}}\frac{1}{(x_{34}^2)^{\omega-2}}\tag{3}$$

If I set $\omega=2$, the integrals decouple and are elementary integrals. This maybe shows why my question arised in the first place.


The important quantities in dimensional regularization are precisely the poles you will obtain in the limit $\omega \rightarrow 2$ and their associated residues. In other words, your bare correlation functions will involve integrals with some divergences, $$ I = \sum_{n = 1}^m\frac{a_n}{(\omega - 2)^n} + \mathrm{finite}, $$ and the renormalization of your theory consists of getting rid of those poles, and in order to do so, you'll need to specify the constants $a_n$.

Now the issue is that, within intermediate calculations, you may have multiple poles contributing. For example, consider the function $$ \frac{f(\omega)}{(\omega - 2)^2}, $$ where $f(2)$ is finite. The problem here is that when you set $\omega$ equal to $2$ within this function, you are actually missing out on a first-order pole in $\omega$. Instead, you should write $$ \frac{f(\omega)}{(\omega - 2)^2} = \frac{f(2)}{(\omega - 2)^2} + \frac{f'(2)}{(\omega - 2)} + \mathrm{finite} $$ where $f'(\omega) = df(\omega)/d\omega$.

This is a potential issue in your case, since your integrals are multiplying $\Gamma(\omega - 2)$ which already has a pole, but I don't quite understand your notation (how are the $\tau_i$'s and $x_{ij}$'s related?). But if the integrals are all finite for $\omega = 2$ then you are safe with the replacement.

  • $\begingroup$ Dear Seth, thank you so much for a very nice answer. I have added details about the notation in the body of the question, but I think I am fine setting $\omega=2$ in the remaining integral. $\endgroup$ – Jxx Oct 9 at 19:10
  • $\begingroup$ I mean, if I were to set $\omega=2$ wrongfully, wouldn’t anyway a divergence show up while integrating? I would think that, in general, one can always set $\omega=2$ and see if the integral converges that way. And if not, do it again with dimensional regularization. Does that make sense? $\endgroup$ – Jxx Oct 9 at 19:42
  • 1
    $\begingroup$ Yes, if it was really a double-pole in $(\omega-2)$, then you would find a divergence if you set $\omega = 2$ too early, so you're correct that the substitution can be used to check the total divergence. My point was just that one also needs the residue of the order-1 pole in $(\omega-2)$. So for example, if you only set $\omega=2$ in one part of the integrand and found a double pole due to other parts, you would miss part of the residue of $(\omega-2)$ which comes from expanding the part of the integrand where you set $\omega=2$. $\endgroup$ – Seth Whitsitt Oct 9 at 20:26
  • $\begingroup$ Yes I see, that’s a good point! Thanks again! $\endgroup$ – Jxx Oct 9 at 22:32
  • $\begingroup$ Actually, I just thought about another case, in which I think that the $\omega=2$ substitution is particularly unwise: what if, after I extracted a divergence in the form of a factor $\Gamma(\omega-2)$, I have an integral that gives $0$ at $\omega=2$? As far as I know, I cannot say that the whole thing is zero, since $0\cdot\infty$ is an indeterminate form. If I encounter that case, does that mean that this is also an instance in which setting $\omega=2$ should not be done? $\endgroup$ – Jxx Oct 10 at 19:36

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