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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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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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