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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.

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    $\begingroup$ For me, your question is not clear. You have the integral with IR & UV divergence. What do you want to demonstrate or find? How UV divergence becomes IR? $\endgroup$ Feb 19 '20 at 17:36
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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.

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  • $\begingroup$ Hey thanks, and apologies for the late reply. Ok, I understand what you mean and this is actually a neat alternative way to see that scaleless integrals are zero. But it doesn't explain to me the deeper argument/why it is "ok" to set $\epsilon_{UV} \equiv \epsilon_{IR}$. What I would like to see is a detailed (if possible) explanation/way of proof for the argument "by analytical continuation I set them equal". $\endgroup$
    – hal
    Jul 24 '20 at 8:18
  • $\begingroup$ I have never seen it in those terms. I am actually not sure I understand what are $\epsilon_{\mathrm{UV}}$ and $\epsilon_{\mathrm{IR}}$. Maybe someone else will be able to help. $\endgroup$
    – MannyC
    Jul 24 '20 at 8:45
  • $\begingroup$ Suppose you have $\int \frac{d^d k_E}{k_E^4}$ with $k_E$ the Euclidian momenta and $d=4- 2\epsilon$ dimensions. Then $\propto \int d k\, k^{-1-\epsilon}$. Now this integral is zero and I can show that by splitting the integration range at an arbitrary point $\int_{-\infty}^{\Lambda}d k\,k^{-1-\epsilon}+\int^{\infty}_{\Lambda} d k\,k^{-1-\epsilon}$. In order to regularize both UV and IR divergences, I take for the first integral $\epsilon \equiv \epsilon_{IR}<0$ and for the second $\epsilon\equiv \epsilon_{UV}>0$. Then the result is $\propto\tfrac{1}{\epsilon_{IR}}-\tfrac{1}{\epsilon_{UV}}$. $\endgroup$
    – hal
    Jul 24 '20 at 9:13
  • $\begingroup$ The point I think it's that you can analytically continue a function from a region where you know it to any other region. You know the first piece for $\epsilon<0$ and the second for $\epsilon>0$. You evaluate those expressions, and then you analytically continue both of them to $\epsilon \in \mathbb{C}$. In dim-reg there is only one $\epsilon$ at all times (at least the way I understand it). $\endgroup$
    – MannyC
    Jul 24 '20 at 9:26

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