In quantum field theory a common term that arises for which we require an expansion is,

$$\frac{\Gamma(2-d/2)}{(4\pi)^{d/2}}\left(\frac{1}{\Delta} \right)^{2-d/2} = \frac{2}{\epsilon}-\log \Delta - \gamma + \log 4\pi + \mathcal O (\epsilon)$$

where $d= 4-\epsilon$, and we supposedly take the expansion about $\epsilon = 0$. Now, when I have done this in Mathematica, I have found terms such as $\sim 2\log \Delta \log 4\pi$ and these have not been included.

In the end, this expansion is used in a calculation in which the $2/\epsilon$ is removed, and $\epsilon\to0$, leaving only the finite part. So, since something like $2\log \Delta \log 4\pi$ is certainly finite and non-zero as $\epsilon \to 0$, I do not see why it has been omitted in this expansion of the expression.

There's also for example a $2\gamma \log 4\pi$. Sometimes constants like these are re-absorbed into redefinitions of what is known as the renormalisation scale in the context wherein these expansions are used in physics, but then I would say why include $-\gamma + \log 4\pi$ and not $2\gamma \log 4\pi$?

I have pondered whether to include this on the physics SE, but as a high ranked frequent user there, I think it would be more appropriate here as I do not believe the answer is physically motivated.

Note: here $\Delta$ can be assumed to be a polynomial in $x$, and that this expansion is to be used in $\int_0^1 dx$. (I wouldn't write out the whole thing here as I do not think it is necessary and scattering amplitudes are messy.)

I get two different expressions depending on whether I apply an asymptotic expansion to each individual part, and then take the product, or I plug the whole thing into Mathematica and compute the series about $\epsilon = 0$:

$$=\frac{1}{8\pi^2 \epsilon} + \frac{1}{16\pi^2}(-\gamma + 2\log 2 + \log \pi -\log \Delta) + \mathcal{O}(\epsilon)$$ $$=\frac{1}{8\pi^2\epsilon} - \frac{\gamma}{16\pi^2} + \frac{1}{16\pi^2}(\log 4\pi - \log \Delta) + \mathcal{O}(\epsilon).$$

  • $\begingroup$ I am sorry if this is trivial, but what is $\Delta$? $\endgroup$ Commented Nov 28, 2016 at 23:56
  • $\begingroup$ @SimpleArt $\Delta$ can vary depending on what we're computing, but in most cases it is a polynomial in $x$, and this expansion is plugged into an $\int_0^1 dx$. $\endgroup$
    – JPhy
    Commented Nov 28, 2016 at 23:57
  • $\begingroup$ Recheck your calculation, you are missing a $\epsilon$ in your cross term proportional to $\log \Delta\log(4\pi)$ (and a overall scaling factor $\frac{1}{16\pi^2}$). $\endgroup$ Commented Nov 29, 2016 at 0:12

2 Answers 2


Let $d/2=2-\epsilon/2$. Then, we can expand the terms of interest as

$$\begin{align} \Gamma(2-d/2)&=\frac{2}{\epsilon}-\gamma +O(\epsilon)\\\\ (4\pi)^{-d/2}&=\frac{1}{(4\pi)^2}e^{\epsilon\log(4\pi)/2}=\frac{1}{(4\pi) ^2}+\left(\frac{\log(4\pi)}{32\pi^2}\right)\,\epsilon+O(\epsilon^2)\\\\ \Delta^{d/2-2}&=\Delta^{-\epsilon/2}=e^{-\epsilon\log(\Delta)/2}=1-\left(\frac{\log(\Delta)}{2}\right)\,\epsilon+O(\epsilon^2) \end{align}$$

Putting it all together yields

$$\begin{align} \frac{\Gamma(2-d/2)}{(4\pi)^{d/2}}\left(\frac1\Delta\right)^{2-d/2}&=\left(\frac{2}{\epsilon}-\gamma +O(\epsilon)\right)\\\\ &\times\left(\frac{1}{(4\pi) ^2}+\left(\frac{\log(4\pi)}{32\pi^2}\right)\,\epsilon+O(\epsilon^2)\right)\\\\ &\times\left(1-\left(\frac{\log(\Delta)}{2}\right)\,\epsilon+O(\epsilon^2) \right)\\\\ &=\frac{1}{(8\pi ^2)\epsilon}-\frac{\gamma}{16\pi^2}+\frac{\log(4\pi)}{16\pi^2}-\frac{\log(\Delta)}{16\pi^2}+O(\epsilon) \end{align}$$

  • $\begingroup$ Your computation is incorrect: you have to carry higher order terms in the expansions of $(4\pi)^{-d/2}$ and $\Delta^{d/2 -2}$, because you are multiplying by the expansion for $\Gamma$, which as a $1/\epsilon$-term. $\endgroup$
    – tracing
    Commented Nov 29, 2016 at 2:13
  • $\begingroup$ @tracing Thank you for the good catch. I've edited. -Mark $\endgroup$
    – Mark Viola
    Commented Nov 29, 2016 at 3:42
  • $\begingroup$ Mark, it seems like there's still a $1/16\pi^2$ factor missing from the $\log \Delta$ term, right? $\endgroup$
    – tracing
    Commented Nov 29, 2016 at 12:19
  • $\begingroup$ A well deserved (+1) $\endgroup$
    – tired
    Commented Nov 29, 2016 at 13:08
  • $\begingroup$ @tracing Yes there is. This is the result of trying to do this in my head. I'll edit - again. Thanks for the heads up! $\endgroup$
    – Mark Viola
    Commented Nov 29, 2016 at 14:55

It seems that the asymptotic formula written in the OP is correct, up to an overall factor of $1/16\pi^2$ (as achille hui notes in comments).

In particular, the cross terms alluded to involving $\log \Delta \log (4\pi)$ have a factor of $\epsilon$ in them, and so can be absorbed into the $O(\epsilon)$ term.


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