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I have recently encountered a truely terrible integral which I need to compute. I am not sure it's doable but before throwing the whole project in the bin I thought I would ask here. At the moment, a step I require is evaluating the integral of $f(z)$ along a closed coutour $C$ containing zero where $f(z)$ is something like

$f(z)=e^{(\frac{a}{z}+\frac{b}{z-c})}g(z)$

$c$ is located outside the contour $g(z)$ is holomorphic inside the disk enclosed by $C$ which has a very long but finite taylor expansion. The reason no traditional tricks work (using the coordinate change $z=1/w$, looking at the series and trying to collect together all the $\frac{1}{z}$ terms) is of course this $\exp((\frac{b}{z-c}))$ term, for which the series expansion has an infinite number of terms, so trying to sum up all those containing $1/z$ is doomed from the start... Has anybody ever encountered something similar? I tried reading about Bessel functions, but they didn't quite fix the problem.

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2 Answers 2

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You can use the residue theorem, given the series expansion of $$h(z)=e^{b/(z-c)}g(z)=\sum_{n=0}^\infty h_n z^n,$$ the contour integral (with $0$ inside and $c$ outside of the contour $C$) evaluates to $$\oint_C e^{a/z}e^{b/(z-c)}g(z)\,dz=2\pi i\sum_{n=1}^\infty \frac{ h_{n-1}a^n}{n!}=2\pi i\sum_{n=1}^\infty \frac{ a^n}{n!(n-1)!}h^{(n-1)}(0),$$ with $h^{(n)}(0)$ the $n$-fold derivative of $h(z)$ evaluated at $z=0$.

I would love to be shown wrong, but I'm pretty certain this is the best one can do in the general case $-$ there is no short-cut to the residue at an essential singularity.

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  • $\begingroup$ Computing the value $c_m$ of the integral with $g:=x^m$ for all $m$ would allow a similar power series approach, for $g=\sum_{m=0}^\infty g_mx^m$ would give $\sum_{m=0}^\infty g_mc_m $. $\endgroup$
    – Pietro Majer
    Commented Jun 2, 2019 at 19:32
  • $\begingroup$ thanks, @PietroMajer , but isn't $c_m=2\pi i a^{m+1}/(m+1)!$ ? so that is not just a similar series but the same series, or did I misunderstand you? $\endgroup$
    – Carlo Beenakker
    Commented Jun 2, 2019 at 19:40
  • $\begingroup$ I'm actually surprised that even the "simplest" case, the unit-circle contour integral of $e^{1/z}e^{1/(z-2)}$, does not seem to have a closed form expression. In that case the series converges rapidly, in 10 terms the same answer as a direct numerical integration is reached (3.26927). $\endgroup$
    – Carlo Beenakker
    Commented Jun 2, 2019 at 21:21
  • $\begingroup$ I'm just saying to write $$\oint_C e^{a/z}e^{b/(z-c)}\Big(\sum_{n=0}^\infty g_mz^m\Big)\,dz=\sum_{n=0}^\infty g_m\Big(\oint_C e^{a/z}e^{b/(z-c)}z^mdz\Big).$$ $\endgroup$
    – Pietro Majer
    Commented Jun 2, 2019 at 22:15
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Your question is too general because you don't specify $g(z)$ and $C$. Something can be done for more concrete cases e.g. $g(z):=z^n,\, n \in \mathbb{N}, C:=\{z:|z|=2\}$ with Maple 2019.1:

int(I*eval(exp(1/z) * exp(2/(z - 1)), z = 2*exp(t*I))*2*exp(t*I), t = 0 .. 2*Pi,numeric);

$ { 1.788426568\times 10^{-14}}+ 18.84955592\,i$

identify(18.84955592*I);

$6\,i\pi $

I think the same can be done with Mathematica.

Addition. Making use of the residue at infinity and Maple, one obtains a simple symbolic result

residue(exp(1/z)*exp(2/(z - 1)), z = infinity + infinity*I)

$-3\pi i$

Therefore, the above integral equals $6\pi i$. Under some conditions on $g(z)$ this should work in the general case too.

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  • $\begingroup$ Yes, I was wondering if there was a general "trick" because this is something that the usual "tricks" fail on. Because my g(z) is holomorphic, I can always taylor series it and reduce it to the sum of your above cases. But it looks like you inputted the sum of exponentials and not the product, if I'm not wrong? If so, this would be a solution to a considerably easier problem. $\endgroup$
    – R Mary
    Commented May 31, 2019 at 18:48
  • $\begingroup$ @R Mary; Thank you. I edited my answer. $\endgroup$
    – user64494
    Commented May 31, 2019 at 18:54
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    $\begingroup$ question: the residue at infinity of $e^{a/z+b/(z-c)}$ is indeed easy to calculate (it's just $-a-b$), but if we transform the contour around the origin into a contour around infinity we would pick up the essential singularity at $z=c$, and so we would need the residue at $c$, which seems to be as difficult to calculate as the residue at $0$. $\endgroup$
    – Carlo Beenakker
    Commented Jun 1, 2019 at 12:13
  • $\begingroup$ @CarloBeenakker: In this case the integral can also be evaluated numerically. $\endgroup$
    – user64494
    Commented Jun 1, 2019 at 15:15

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