I want to find the residues of the integral

$F = \int_{-\infty}^{\infty} \dfrac{1}{x+(a-ib)} \dfrac{1}{\exp(-x/c)-1} dx$

I know that $x=-(a-ib)$ is a simple pole which contributes a non-zero residue.

How can I use the Laurent series expansion of $\dfrac{1}{\exp(-x/c)-1}$ to determine if $x=2\pi icn$, $n = 0, \pm 1, \pm 2, ...$ are also simple poles which give non-zero residues?

(The Laurent expansion at $x=0$ is $-\dfrac{c}{x} - \dfrac{1}{2} - \dfrac{x}{12c} + \dfrac{x^3}{720c^3} + O(5)$ )

  • $\begingroup$ First of all, one does not find residues of an integral, but rather residues of poles of a function that is an integrand of an integral. Second, this particular integral does not converge due to the pole at $x=0$. $\endgroup$
    – Ron Gordon
    Dec 18, 2015 at 7:17
  • $\begingroup$ Good point, @RonGordon. Could you further explain the non-convergence issue? Is the $x=0$ a pole that I use to calculate a residue? $\endgroup$ Dec 18, 2015 at 7:37
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    $\begingroup$ The pole lies on the contour, so no. However, if you want to compute the Cauchy principal value of the integral, which does exist, then you may put a small semicircular detour around the pole at $z=0$ - this is a standard technique. If the detour goes above the pole, then you simply compute the integral about the contour, which usually is equivalent to half a residue of the pole there. But first you need to decide if it is the Cauchy PV of the integral that you want. $\endgroup$
    – Ron Gordon
    Dec 18, 2015 at 7:45
  • $\begingroup$ That sounds reasonable. Could you write this as an answer below and if possible, include some good references to what you said? $\endgroup$ Dec 18, 2015 at 7:51
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    $\begingroup$ I would, but now that I look at the integral further, even the Cauchy PV does't seem to converge either. Depending on the sign of $c$, the integrand converges at one of the endpoints as $1/x$, which is non-integrable. $\endgroup$
    – Ron Gordon
    Dec 18, 2015 at 13:59

1 Answer 1


Consider $\,x:=2\pi icn+y\;$ with $n\in \mathbb{Z}\,$ then $$\frac{1}{\exp(-x/c)-1}=\frac{1}{\exp(-2\pi in-y/c)-1}=\frac{1}{\exp(-y/c)-1}$$ Since for $\,x\to 2\pi icn\;$ we have $\,y\to 0\;$ your expansion may be used with $y$ instead of $x$.
Replacing $\,y\,$ by $\;(x-2\pi icn)\;$ there will give you the Laurent expansion as $\,x\to 2\pi icn$.

The residues of $f(x):=\dfrac{1}{x+(a-ib)} \dfrac{1}{\exp(-x/c)-1}$ at $\,x= 2\pi icn\,$ will then be given by $\;\operatorname{Res}(f(x),2\pi icn)=\dfrac{1}{2\pi icn+(a-ib)}(-c)\ \ $ (at least if $\,a\neq 0\,$ or $\,2\pi cn\neq b$).

Hoping this clarified more (else ask questions!).

  • $\begingroup$ Since I defined $x=2\pi icn$ where $n=0,\pm 1,\pm 2,...$, shouldn't the residue expression have a summation, $\sum_n$? $\endgroup$ Dec 18, 2015 at 7:57
  • $\begingroup$ @MedullaOblongata: yes of course you will have to add all the residues in your contour (and not forget the residue at $-(a-ib)$). $\endgroup$ Dec 18, 2015 at 8:31

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