The residue theorem is a standard result in complex analysis, I state it below so we are on the same page: note that $\overline{\mathbb{C}}$ is the extended complex plane (ie. $\simeq$ Riemann sphere)

Let $f : \overline{\mathbb{C}} \rightarrow \overline{\mathbb{C}}$ be a function which is analytic inside a positively-oriented closed contour $C \subset \overline{\mathbb{C}}$ apart from at a finite number of singular points $a_{1}, \ldots a_{n}$ which are contained inside $C$. Then we have the following result known as the residue theorem, \begin{eqnarray} \int_{C} f(z) \ dz = 2 \pi i \sum_{\nu = 1}^{n} \text{Res}_{z=a_{\nu}}f(z). \quad \quad \quad \quad \quad (1)\end{eqnarray}

My question is: what if $n = \infty$?

There must be extra conditions in this case because, while the left hand side of (1) may converge, the right hand side of (1) becomes a series and may not converge.

A thought: if $\sum_{\nu=1}^{\infty} \text{Res}_{z=a_{\nu}}f(z)$ is only a formal series, then perhaps it represents $\frac{1}{2\pi i}\int_{C}f(z) \ dz$ asymptotically. ie. $\frac{1}{2\pi i}\int_{C}f(z) \ dz \sim \sum_{\nu=1}^{n} \text{Res}_{z=a_{\nu}}f(z)$ as $n \rightarrow \infty$, is this the answer?


1 Answer 1


I think that is hardly possible. If $\overline C$ is a compact set then the set of poles for $f$ in $\overline C$ will have a limit point $a$ and at that point $f$ will have a essential singularity. This complicates it.

The proof of the theorem depends on you being able to define the residue for every point in $C$, but that's not possible for the singularity at $a$.

You could of course do something similar to indefinite integral where you take limes as the curve encloses all poles:

$\lim_{n\to\infty}\int_{C_n} f(z) dz = \lim_{n\to\infty}2\pi i \sum \operatorname{Res}_{a_k}f$

But where the expresson under the right limes is of finitely many terms.

  • $\begingroup$ What about $\int _{- i \infty}^{+ i \infty} \Gamma(z+1) \ dz$, this has infinitely many poles $\endgroup$
    – analytic
    Commented Sep 21, 2015 at 13:15
  • $\begingroup$ But is $\Gamma$ differentiable at $\infty$? $\endgroup$
    – skyking
    Commented Sep 21, 2015 at 13:25
  • $\begingroup$ The path of integration could be deformed so that the point at $\infty$ is avoided, but still infinitely many poles are contained in the contour. Is this right? $\endgroup$
    – analytic
    Commented Sep 21, 2015 at 13:32
  • $\begingroup$ If you avoid $\infty$ then the integration path is deformed in a way that you only will enclose finitely many poles. $\endgroup$
    – skyking
    Commented Sep 21, 2015 at 13:34
  • $\begingroup$ The singularities of $\Gamma(z+1)$ are at $z = -1,-2,-3,\ldots$ i.e. all integers along the negative real axis. What about if I deform the path of integration by making a semi-circle crossing the positive real axis, thus avoiding $\infty$ and keeping infinitely many singularities inside the contour? Hopefully that's clear $\endgroup$
    – analytic
    Commented Sep 21, 2015 at 13:37

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