I was reading a paper that gives a nice collection of all scalar integrals that crop up in QCD loop calculations. Such integrals are computed in some kinematic region and then the authors say the results may be analytically continued if so desired. I just wonder how is this analytic continuation done in practice? It's a relatively short paper and the url is https://arxiv.org/pdf/0712.1851.pdf

The authors state that a particular kinematic region allows for the $i\epsilon$ to be dropped and then one can analytically continue results via the prescription $p_i^2 \rightarrow p_i^2 + i\varepsilon, s_{ij} \rightarrow s_{ij} + i\varepsilon, m_i \rightarrow m_i - i\varepsilon$. I just wonder why this is the case and if the sign choices here are significant?

As a simple example, the analytic continuation of the massive tadpole is given as $I_1^D(m^2) = -\mu^{2\epsilon} \Gamma(\epsilon-1) (m^2-i\varepsilon)^{1-\epsilon}$ but what should I do with this result as it contains an explicit $\varepsilon$ in $m^2-i\varepsilon$?

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    $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/0712.1851 $\endgroup$ – Qmechanic Sep 13 '17 at 19:27

This is significant: they are originate from the $+i\varepsilon$ prescription in the Feynman propagators of the original integral. The sign choices are important to obtain the correct sign for the imaginary part of the loop functions.

Note that these prescriptions are relevant only if evaluations are being made on a branch cut. Therefore, for your example, if $m^2 > 0$ (which is usually the case), the $-i\varepsilon$ can be dropped.

You might find my computer program $Package$-X for Mathematica very helpful, since it produces all loop functions in any kinematic configuration consistent with the $i\varepsilon$ prescription. And they are ready for numerical evaluation directly within Mathematica.

  • $\begingroup$ Thanks for your answer! In fact, I have already been using PackageX extensively. In, for example, LoopIntegrate[1,k,{k,0},{k+p1,mc},{k+p1+p2,mc}] together with scalar product relations, the package will sometimes just return an expression without me specifying the domain of parameters. In this case, would the expression be valid in the parameter region in which there are no imaginary parts? And how would one in general, through PackageX, specify the parameter region one wants the expression in? Thanks! $\endgroup$ – CAF Jul 22 '18 at 9:21
  • $\begingroup$ @CAF Ah I missed your comment, sorry. The expression returned by Package-X is supposed to be valid for all real values of the scalar products and all positive values for the internal masses. Sometimes, results are expressed in terms of Ln and DiLog to accommodate the $+i\varepsilon$ prescription over the whole domain of reals. But, if you know that some quantities are restricted to a smaller domain, you can wrap Assuming around LoopRefine (or C0Expand, D0Expand, etc), and that will make it try to come up with something in terms of Mathematica's Log and PolyLog only... $\endgroup$ – HirenPatel Jul 25 '18 at 4:39
  • $\begingroup$ ...for example you can do something like Assuming[s<4m^2, ExpandC0[...]] and then it will try to come up with an explicit form of ScalarC0 with fewer Ln and DiLog. This part of Package-X definitely needs more work; in a future version, I'll provide the option Assumptions for LoopRefine, C0Expand, etc., and improve the code to produce better results in cases when some variables are restricted to a smaller domain. $\endgroup$ – HirenPatel Jul 25 '18 at 4:46
  • $\begingroup$ Thanks for your reply. As an example, I tried the following Assuming[{xb > 1, x < -4*mc^2, mc > 0}, C0Expand[ScalarC0[mc^2, x/4 + mc^2/xb + x/(4 xb), 2 mc^2 + x/2 + (2 mc^2)/xb + x/(2 xb), mc, 0, mc]]]. There is no simplification made from DiLog to PolyLog that I can see. Does the Assuming only work for relatively simpler ScalarC0 functions? $\endgroup$ – CAF Jul 26 '18 at 10:10
  • $\begingroup$ In any case, it doesn't take long to analyse term by term to check whether the DiLog is a PolyLog or a Conjugate[PolyLog[]]. The PolyLog[2,x] is simply $\text{Li}_2(x)$. Is there a rewriting of Conjugate[PolyLog[2,x]]? $\endgroup$ – CAF Jul 26 '18 at 10:12

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