I suppose a stupid question but I was wondering about it for a while:

Can one apply the residue theorem to a function $f$ which is defined and holomorphic on $U-\{a_1,a_2,\dots\}$ where $U$ is simply connected open subset of the complex plane and $a_k$ for $k\geq 1$ are all simple poles of $f(z)$. In particular I am thinking about the case when $f(z)= \Gamma(z)$, the gamma function. It is known that it has simple poles at $z=-k$, $k=0,1,2,\dots$ with resdue $(-1)^k/k!$.

I hope the question is clear. Excuse me in case it is to trivial. I am not an expert in complex analysis.

The Wikipedia-article demands a finite number of points $a_k$. Will one get a problem with a suitable choosen closed contour?

To be more precise I would like to calculate the following integral:

$$\int_{-\infty}^{\infty} dx f(x)$$

I was wondering if the typical "trick" of constructing a closed half-circle in the upper-half plane would work too when the poles continue ad infinitum.


My initial motivation is to invert a Mellin transform using the Mellin inversion theorem, when the Mellin transform has an infinite number of isolated poles.

For example: I am able to show (by using a specific symmetry of my problem) that the Melin transform of a function $P(x)$ for $0\leq x\leq x_c$ fulfills the following equation:

$$M(s) = \frac{2x_c^s}{s-2+x_c^s}$$

where $0<x_c<1$. Now when I try to apply the Mellin inversion theorem I need to know where the poles lie. Unfortunately I think there is an infinite number poles of $M(s)$.

  • $\begingroup$ Can you imagine a closed contour encircling all of the poles of $\Gamma(z)$? $\endgroup$ – Antonio Vargas Jun 20 '14 at 16:46
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    $\begingroup$ No, not really in case of $a_k=-k$ for $k=0,1,\dots$. But it should be possible to find a contour when I can bound the $a_k$'s, for example when $a_k=1/k$ for $k=1,2,\dots$. In this a contour can be constructed... But are there any other technical problems with the residue theorem in this case... $\endgroup$ – antarcticfox Jun 20 '14 at 17:02
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    $\begingroup$ The complication in the case where there are infinitely many poles in a compact region is that you end up with a point where the singularity is not isolated and there may be no good way to circumvent that point in order to approximate by a contour enclosing only a finite number of poles. $\endgroup$ – DisintegratingByParts Jun 21 '14 at 2:16
  • $\begingroup$ Alright, you get a problem with non-isolated poles, but when you have an infinite sequence of isolated poles, like $a_k=-k$. An answer in a similar post (math.stackexchange.com/questions/205057/…) states: "To summarize, you can apply the residue theorem to an "infinite contour" enclosing infinitely many singularities, as long as any of the actual contours in the sequence justifying this argument enclose only finitely many singularities; otherwise you have a non-isolated singularity to which the residue theorem doesn't apply." $\endgroup$ – antarcticfox Jun 21 '14 at 7:25
  • $\begingroup$ So I suppose it can work in this case to apply the Mellin inversion theorem as long as $M(s)$ the function with infinitely many poles has the property of tending to zero uniformly when $Im(s) \rightarrow \infty$, c.f. en.wikipedia.org/wiki/Mellin_inversion_theorem, not sure though... $\endgroup$ – antarcticfox Jun 21 '14 at 7:31

It does work for an infinite number of singularities, as long as they are all isolated.

From Ahlfors' Complex Analysis, p.150:

$$\frac{1}{2\pi i} \int _\gamma f \ dz = \sum_{j} n(\gamma, a_j) R_j$$ This is the residue theorem, except for the restrictive assumption that there are only a finite number of singularities. In the general case, we need only prove that $n(\gamma, a_j) = 0$ except for a finite number of points $a_j$. The assertion follows by routine reasoning. The set of points $a$ with $n(\gamma, a) = 0$ is open and contains all points outside a large circle. The complement is consequently a compact set, and as such it cannot contain more than a finite number of isolated points $a_j$. Therefore $a_j \ne 0$ for a finite number of the singularities.

  • $\begingroup$ Excuse me. The complement is a compact set, but what's wrong? Compact subset of complex plane is limit point compact since the complex plane is metric space, so if the set of isolated points is infinite that set must have a accumulation point, but this does not mean that accumulation point should be one of the isolated singularities.. doesn't it? $\endgroup$ – Guldam Jun 11 '15 at 10:53
  • $\begingroup$ I think @Guldam is right. The closed interval $[0,1]$ is a compact set which contains infinitely many isolated points (e.g., $1/n$ for $n \in \mathbb{N}$), contrary to Ahlfors' statement "a compact set contains only finitely many isolated points". $\endgroup$ – Randy Randerson Nov 11 '15 at 6:55
  • $\begingroup$ @Guldam - I believe that Ahlfors is trying to say there are only finitely many isolated singularities whose winding numbers are non-zero. He showed that if any point is outside a closed disk, then the winding number is zero. So, the set of isolated singularities with non-zero winding number is inside the compact disk. If there were infinitely many, then there would be an accumulation point in the disk for the set of isolated singularities. There's no way for the function to be differentiable at this accumulation point, so it is a singularity, but is not isolated, contrary to assumption. $\endgroup$ – bryanj Nov 14 '15 at 18:49
  • $\begingroup$ The set of singularities has to be closed, Ahlfors doesn't say this explicitly but it's assumed I believe. If you assume this, all of these arguments work, also the holomorphic function $f$ in the residue theorem is defined in a region \ set of singularities, this set has to be open to be able to talk about holomorphism, this implies that the set of singularities is closed. Also the Wikipedia article on isolated singularities says that cluster points of isolated singularities counts as "singular behavior". By closed I mean closed In the region under consideration. $\endgroup$ – Zero Feb 23 '18 at 1:35
  • $\begingroup$ I made a slight mistake in my previous comment by closed I mean closed in the whole plane, not just the region. Because otherwise we could take the plane \ 0, the set of singularities as 1/n where n is natural and the curve the unit circle. $\endgroup$ – Zero Feb 23 '18 at 4:02

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