Analytical continuation in QFT My question is quite basic and generic. It is known that scaleless integrals that appear in QFT such as
$\int \frac{d^dk}{(k^2)^2} = \frac{1}{\epsilon_{\mathrm{UV}}} -
\frac{1}{\epsilon_{\mathrm{IR}}} = 0$ in $d=4-2\epsilon$ dimensions,
by analytically continuing $\epsilon_{\mathrm{UV}} \rightarrow \epsilon_{\mathrm{IR}}$. I have read about the proper definition from mathematical point of view, of how the analytic continuation works but I fail to see how after it, both $\epsilon$'s are equal. 
 A: Dimensional regularization comes with an additional prescription for analitically continuing divergent integrals that do not have a domain of convergence in the complex $D$ plane. Namely, if an integral can be written as the sum con a finite number of integrals which have mutually disconnected domains of convergence, then the integral is defined to be the sum of all those contributions.
Consider now
$$
I_\alpha := \int \frac{\mathrm{d}^Dp}{(2\pi)^D} \frac{1}{(p^2)^\alpha}\,.
$$
Let us do the simpler case first: $\alpha = 0$. 
Then one can do
$$
I_0 = \int \frac{\mathrm{d}^Dp}{(2\pi)^D} \frac{m^2}{p^2+m^2} + \int \frac{\mathrm{d}^Dp}{(2\pi)^D} \frac{p^2}{p^2+m^2} := I_0^{(1)} + I_0^{(2)}\,.
$$
having splitted $\frac{p^2+m^2}{p^2+m^2} = 1$ into two summands. It is easy to see that the domain of convergence $\mathcal{D}_k$ of $I_0^{(k)}$ is given by
$$
\begin{aligned}
\mathcal{D}_1 &= \{D \in \mathbb{C}\,:\,0 < \mathfrak{Re}\, D < 2\}\,,\\
\mathcal{D}_2 &= \{D \in \mathbb{C}\,:\,-2 < \mathfrak{Re}\, D < 0\}\,.
\end{aligned}
$$
So their domains of convergence do not overlap. We have to use the prescription stated above, so we evaluate both integrals and then sum them. It turns out
$$
\begin{aligned}
I_0^{(1)} &= m^D \frac{\Gamma\left(1-\frac{D}2\right)}{(4\pi)^{D/2}}\,,\\
I_0^{(2)} &= m^D\frac{D\,\Gamma\left(-\frac{D}2\right)}{2(4\pi)^{D/2}}\,.\\
\end{aligned}
$$
After using $\Gamma(x+1) = x\, \Gamma(x)$ one can see that the sum of those contributions vanishes.
Now, what about the general case? One can always do the same splitting
$$
\begin{aligned}
I_\alpha &= \int \frac{\mathrm{d}^Dp}{(2\pi)^D} \frac{1}{(p^2)^\alpha} \frac{p^2+m^2}{p^2+m^2}  \\
&= \int\frac{\mathrm{d}^Dp}{(2\pi)^D} \frac{m^2(p^2)^{-\alpha}}{p^2+m^2}+\int\frac{\mathrm{d}^Dp}{(2\pi)^D} \frac{(p^2)^{1-\alpha}}{p^2+m^2} := I_\alpha^{(1)} + I_\alpha^{(2)}
\,.
\end{aligned}
$$
Now the integral $I_\alpha^{(k)}$ has domain of convergence
$$
\mathcal{D}_k = \{D \in \mathbb{C} \,:\, 2\alpha +2- 2k < \mathfrak{Re}\,D < 2\alpha +4- 2k\}\,.
$$
They are also disjoint. Now the integrals are almost identical to the previous ones
$$
\begin{aligned}
I_\alpha^{(1)} &= m^{D-2\alpha} \frac{\Gamma\left(1+\alpha-\frac{D}2\right)\Gamma\left(-\alpha+\frac{D}2\right)}{(4\pi)^{D/2}\Gamma\left(\frac{D}2\right)}\,,\\
I_\alpha^{(2)} &=  m^{D-2\alpha} \frac{\Gamma\left(1-\alpha+\frac{D}2\right)\Gamma\left(\alpha-\frac{D}2\right)}{(4\pi)^{D/2}\Gamma\left(\frac{D}2\right)}\,.\\
\end{aligned}
$$
Using the same property as before
$$
I^{(1)}_\alpha + I^{(2)}_\alpha = 0\,.
$$
In general, if one has a function $f(p^2)$ which has a behavior $f(p^2) \sim (p^2)^{\alpha_{\mathrm{UV}}}$ in the UV and $f(p^2) \sim (p^2)^{\alpha_{\mathrm{IR}}}$ in the IR it is still possible to do the same trick. It suffices to multiply by 
$$
\left(\frac{p^2+m^2}{p^2+m^2}\right)^n = 1 = \sum_{k=0}^n \binom{n}{m} \frac{(p^2)^{k}}{(p^2+m^2)^n}\,.
$$
For $n$ big enough (namely $n > \alpha_{\mathrm{UV}} - \alpha_{\mathrm{IR}})$ all domains of convergence are non empty, indeed they are
$$
\mathcal{D}_k = \{D \in \mathbb{C} \,:\, -2\alpha_{\mathrm{IR}} - 2k < \mathfrak{Re}\,D < -2\alpha_{\mathrm{UV}} + 2n- 2k\}\,.
$$

Reference: RENORMALIZATION Damiano Anselmi.
