Consider the function,

$$f(z) = \frac{\sin(\frac{1}{z})}{z^2+a^2} \text{ where } a>0$$

I know that this has two simple poles at $z = \pm ia$ and an essential singularity at $z = 0$ because of $\sin(\frac{1}{z})$ in the numerator. For $\operatorname{Res}(f,ia)$ and $\operatorname{Res}(f,-ia)$, it is relatively easy to find the these values using

$$a_{-1} = \lim_{z\rightarrow z_0} ((z-z_0)f(z))$$

However, I am finding it much more difficult to determine the residue at an essential singularity. From what I have read, it seems that the easiest way to determine the residue is to explicitly find the Laurent Expansion and then get it from there. Would this be the case for this function as well?

  • $\begingroup$ Actually the easiest way is to use the fact that the sum of all residues of $f$ is $0$. $\endgroup$ – Daniel Fischer Sep 30 '17 at 20:08
  • $\begingroup$ How do we know that? I can't seem to find that in any of the reading. Also, I'm finding that $Res(f, -ia) = Res(f, ia) = \frac{sin(\frac{1}{ia})}{2ia}$ and $Res(f, 0) = \frac{1}{a^2}$ which does not add up to zero in this case. $\endgroup$ – Baugh_Mania Sep 30 '17 at 20:15
  • 1
    $\begingroup$ You have the wrong residue at $0$. This is a quite similar question. $\endgroup$ – Daniel Fischer Sep 30 '17 at 20:20
  • $\begingroup$ Alright, so if I'm not mistaken $\oint_C f(z) dz = Res(f, ia)+Res(f, -ia)+Res(f, 0)$ therefore, if we know two of the residues and can solve the contour integral then we can find the third residue. Is that correct? $\endgroup$ – Baugh_Mania Sep 30 '17 at 20:27
  • $\begingroup$ Modulo a factor of $2\pi i$, and the condition that $C$ encloses all three singularities. Take $C$ as $\{ z : \lvert z\rvert = R\}$ for $R > \lvert a\rvert$, and let $R \to +\infty$ to see that the integral is $0$. $\endgroup$ – Daniel Fischer Sep 30 '17 at 20:33

For positive numers $a<r<R$, Cauchy's Integral Theorem guarantees that


Then, noting that $\left|\sin\left(\frac1{Re^{i\phi}}\right)\right|\le \frac{R}{R-1}$ for $R>\max(1,a)$

$$\begin{align} \left|\oint_{|z|=R}\frac{\sin(1/z)}{z^2+a^2}\,dz\right|&=\left|\int_0^{2\pi}\frac{\sin\left(\frac1{Re^{i\phi}}\right)}{R^2e^{i2\phi}+a^2}\,iRe^{i\phi}\,d\phi\right|\\\\ &\le \int_0^{2\pi} \frac{\left|\sin\left(\frac1{Re^{i\phi}}\right)\right|}{\left|R^2e^{i2\phi}+a^2\right|}\,R\,d\phi\\\\ &\le \frac{R^2}{(R^2-a^2)(R-1)}\\\\ &\to 0\,\,\text{as}\,\,R\to \infty \end{align}$$

Hence, the sum of the residues of $\frac{\sin\left(1/z\right)}{z^2+a^2}$ is $0$ and we find that

$$\begin{align} \text{Res}\left(\frac{\sin\left(1/z\right)}{z^2+a^2},z=0\right)&=-\text{Res}\left(\frac{\sin\left(1/z\right)}{z^2+a^2},z=ia\right)-\text{Res}\left(\frac{\sin\left(1/z\right)}{z^2+a^2},z=-ia\right)\\\\ &=\frac{\sinh(1/a)}{a} \end{align}$$

  • $\begingroup$ reversing the contour, pretty nice (+1) $\endgroup$ – tired Oct 1 '17 at 18:45
  • $\begingroup$ @tired Thank you my friend! Nice to see you're still around. How are you? $\endgroup$ – Mark Viola Oct 1 '17 at 22:41
  • $\begingroup$ thanks i'm pretty fine just way too busy to contibute here on MSE regularly :(. How did it go for you the last time? Everything alright? $\endgroup$ – tired Oct 1 '17 at 23:10
  • $\begingroup$ @tired Pleased to hear. I've been lightly contributing here. There aren't many interesting integrals or series anymore. $\endgroup$ – Mark Viola Oct 1 '17 at 23:26
  • $\begingroup$ I have exactly the same feeling..it is pretty sad somehow :( $\endgroup$ – tired Oct 2 '17 at 0:10

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