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Motivation: In my research I try to find the asymptotics of the heat trace $\text{tr}e^{-t\Delta}$ as $t\to0$, where $\Delta$ is Laplace operator on a manifold with singularities. First I find the asymptitics of the trace of the resolvent for some $n\in\mathbb{N}$,i.e. $\text{tr}(\Delta-\mu)^{-n}$ and then using Cauchy formula $\text{tr}e^{-t\Delta}=\frac{(n-1)!}{2\pi i}\int_{\Gamma}e^{-t\mu}\text{tr}(\Delta-\mu)^{-n}d\mu$ find the desired asymptotics. On a compact Riemannian manifold asymptotics of the trace of the resolvent is given by power series so the contour integral is easy to calculate, but on manifold with singularities some logarithmic terms might occur and I do not know how to deal with the corresponding integrals. The exact statement of the problem is the following.

Let $m\in\mathbb{N},t>0$ how to compute the integral $\int_{\gamma}e^{-t\mu}(-\mu)^{-m}\log\sqrt{-\mu}d\mu$, where $\gamma$ is contour $\{|\arg(\mu+1)|=\pi/4\}$ transversed upward?

Here are my thoughts:

If I try to use the Residue theorem I need a closed contour. Assume 0 is in the contour then we have troubles with $\log$ because the singularity at 0 is not a pole! And if 0 is not in the contour the the integral is zero... Is there any contour such that the integral is not zero?

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  • $\begingroup$ First, you need to choose a branch-cut for the multivalued function $\log\sqrt{−\mu}$, say $ℝ_{≥0}$. Then I'd try to obtain your contour as the limit of a compact, closed contour to which you apply the residue formula. For instance the boundary of the domain $(\{|\arg(μ+1)|≤π/4\}∩D(0,r))\setminus(D(0,1/r)∪\{Re(\mu)>0 , |Im(μ)|<1/r\}$ for $r>1$ tending to infinity. But without you giving more research-level background and motivation on this matter, your question is likely to be closed as off-topic. $\endgroup$ Feb 26, 2016 at 17:06
  • $\begingroup$ Thank you for the help. As I understand the integral over this closed contour is zero because the function $e^{-t\mu}(-\mu)^{-m}\log\sqrt{-\mu}$ has no poles inside the domain. Is it true? $\endgroup$
    – Asya
    Feb 27, 2016 at 11:35
  • $\begingroup$ Yes, of course. The point is that the integral on the compact contour is the sum of something tending to your integral as $r\to \infty$ plus something tending to something else (which is then the opposite of the value you're looking for). I'm not going to give further help on this matter as this particular discussion is way below research level (I'm not speaking about the context you provided, just the specifics about complex analysis you're bringing up just now). $\endgroup$ Feb 27, 2016 at 14:04
  • $\begingroup$ Well.. I do not see how the change of the contour you described helps to solve the problem I have no idea how to take the integral along any line. But I think it has connection to Gamma function. As you suggested I added a motivation to the question and also posted the question here but there are no answers. So where can I have a real help with the question? $\endgroup$
    – Asya
    Feb 28, 2016 at 13:22
  • $\begingroup$ I didn't say what I suggested was certain to reach the goal... it was only a general advice. Honestly I can't help you here because I can't invest time on this topic, sorry about that. If you think the result relates to Gamma function then you should try to find a change of variables leading to Henkel's Gamma integral representation. I wish you luck! $\endgroup$ Feb 28, 2016 at 15:07

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