An asymptotic expansion involving the gamma function in QFT 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 $\epsilon = 0$:
$$=\frac{1}{8\pi^2 \epsilon} + \frac{1}{16\pi^2}(-\gamma + 2\log 2 + \log \pi -\log \Delta) + \mathcal{O}(\epsilon)$$
$$=\frac{1}{8\pi^2\epsilon} - \frac{\gamma}{16\pi^2} + \frac{1}{16\pi^2}(\log 4\pi - \log \Delta) + \mathcal{O}(\epsilon).$$
 A: 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
$$\begin{align}
\frac{\Gamma(2-d/2)}{(4\pi)^{d/2}}\left(\frac1\Delta\right)^{2-d/2}&=\left(\frac{2}{\epsilon}-\gamma +O(\epsilon)\right)\\\\
&\times\left(\frac{1}{(4\pi) ^2}+\left(\frac{\log(4\pi)}{32\pi^2}\right)\,\epsilon+O(\epsilon^2)\right)\\\\
&\times\left(1-\left(\frac{\log(\Delta)}{2}\right)\,\epsilon+O(\epsilon^2)
\right)\\\\
&=\frac{1}{(8\pi ^2)\epsilon}-\frac{\gamma}{16\pi^2}+\frac{\log(4\pi)}{16\pi^2}-\frac{\log(\Delta)}{16\pi^2}+O(\epsilon)
\end{align}$$
A: 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.
