The residue theorem is commonly used to calculate situations where the number of isolated singularities is limited. However, I am very curious whether the residue theorem can be extended to cases with infinite singularities.

I hope to find a concise function to explain this conjecture and find a few counterexamples. But it's not easy.

So, I will use an integration problem that I actually encountered as an example to illustrate. Of course, we are not limited to the example given below.

This integration problem was obtained when I attempted to solve a magnetic field problem. It is shown in the blow. $$ \int_{-\infty}^{\infty}{\frac{e^{gx}\cosh\mathrm{(}tx)\mathrm{e}^{jpx}}{x(x+\beta )\left[ a(e^{2gx}+1)+b(e^{2gx}-1) \right]}}\mathrm{d}x \\ where\,\,0<p,0<\beta ,0<a<b,0<t<g $$ In order to solve this inverse Fourier transform problem using residue theorem, let an auxiliary complex variable function: $$f(z)=\frac{e^{gz}\cosh\mathrm{(}tz)\mathrm{e}^{jpz}}{z(z+\beta )\left[ a(e^{2gz}+1)+b(e^{2gz}-1) \right]}$$

Obviously, the first-order pole of $f (z)$ is $0$, $-\beta$, and $$k_n=\frac{j2n\pi +\ln \left( \frac{b-a}{b+a} \right)}{2g}$$. Where, $n\in \mathbb{Z}$.

We can construct an infinitely large semicircular integral contour $C$ in the upper half complex plane. As shown in Fig.1.

Integral contour $C$

Based on Jordan's lemma, it is convenient to calculate the integration of various arc paths on the integral contour C, including infinite diameter semi-circular arcs $C_R$ and small arcs $C_{\varepsilon 1}$$C_{\varepsilon 2}$ and $C_{\varepsilon 3}$ near poles on the x-axis.

And, by calculation, we can confirm that the infinite series $\sum_{n=1}^{\infty}Res[f(z),k_n]$ is convergent.

So, our conjecture is whether $\oint_{C}f(z)dz = 2j\pi\sum_{n=1}^{\infty}Res[f(z),k_n]$ holds true. Will it is universally valid ?

Is this a proven problem? Or can we find a suitable counterexample. I have not found any relevant papers discussing this issue.

  • 3
    $\begingroup$ The application of the residue theorem needs a finite contour, enclosing a finite number of isolated singularities. This applies to your case. Then you can take the limit of the contour to infinity, but each contour in this sequence is finite. (Note that a finite contour cannot have an infinite number of isolated singularities.) See for example math.stackexchange.com/a/1445021/87355 $\endgroup$ Commented Apr 30 at 9:28
  • $\begingroup$ Thanks for your help, Carlo Beenakker. $\endgroup$
    – adios518
    Commented May 19 at 15:27

1 Answer 1


Your function has simple poles at $0,-\beta$ and $k_0$. So strictly speaking your integral diverges. The standard way to bypass this difficulty is to consider it as a principal value. Then the residues at these three simple poles will enter to the final answer, a half of each residue must be included.

Now, to check that "Jordan's lemma" applies here, it is easier to consider a rectangular contour (instead of the big semicircle). And to show that its contribution to the integral tends to zero (this is true in your example but has to be carefully checked).

On your last question: this is not "universally valid". Each particular example requires a separate consideration: the residue theorem is applied to some contours on which your function is analytic, and then you pass to the limit. In your case this involves an estimate of the integral over the big half-circle (or rectangle).

Remark. Don't use the word "countless" in a mathematical text; it is not an accepted term. Mathematicians use "infinite" and "uncountable", but these two words have very different meaning. So for example, the set of zeros of an analytic function of one variable can be infinite but it is always countable.

  • $\begingroup$ Thanks for your help, Alexandre Eremenko $\endgroup$
    – adios518
    Commented May 19 at 15:28

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