By Mittag-Leffler's theorem if we want a function with poles at $z=1$ and $z=2$ we can write $$ f_1(z) = \frac{z}{2}\sum_{k=1}^2\frac{1}{k(z-k)} $$ The sum of the residues is just $1$ because of the factor of $2$. I wanted to extend this to a function with a 'continuous line of poles' between $1$ and $2$, and thought of $$ f_\infty(z) = z\int_1^2\frac{1}{k(z-k)}\;dk = \log(1-z)-\log\left(1-\frac{z}{2} \right) $$ this does have a gap in the requested range, but seems to have defined complex values at $1<z<2$. We could work up to this function using something like $$ f_{\infty}(z)=\lim_{n \to \infty} f_n(z)=\lim_{n \to \infty}\frac{z}{n+1}\sum_{k=0}^n \frac{1}{\left(1+\frac{k}{n}\right)\left(z-1-\frac{k}{n} \right)} $$ slowly filling in more and more poles. Because of the factor of $n+1$, the sum of the residues of the right hand side is still just $1$ for any $n$. The function appears to converge to $f_{\infty}(z)$ (outside the gap) when plotting.

Does this mean we can say the sum of the residues of $\log(1-z)-\log\left(1-\frac{z}{2} \right)$ is also $1$ in some limiting sense?

Are there any functions that truly have a 'continuous line of poles' in some sense and have well defined residues?

Thanks for any input.

  • $\begingroup$ What do you mean by "continuous line of poles" $\endgroup$ – Ben P. Nov 15 '17 at 10:59
  • $\begingroup$ @BenP. Thanks for your comment. Whatever is could mean. Hence why it's in quotes. A function that is infinity in a small region that's bigger than one point. If that's impossible or doesn't make sense I'd like to know why. I picked that way of talking about it because I was considering building the poles up step by step by taking the limit $n \to \infty$. $\endgroup$ – Benedict W. J. Irwin Nov 15 '17 at 11:04
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    $\begingroup$ Like the zeroes of an analytic function, the poles of a meromorphic function cannot accumulate. $\endgroup$ – Jack D'Aurizio Nov 15 '17 at 11:58
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    $\begingroup$ You mean a branch cut? That's what you have $\endgroup$ – Dylan Nov 15 '17 at 19:48
  • $\begingroup$ @JackD'Aurizio and Dylan Thank you both for your information. I think branch cut is the term I was looking for here and I will read up about this. So there is a sense that tightly packed poles can morph into a branch cut (as based on the question content)? That is, if such a conversion was significantly different to the accumulation that Jack has stated is forbidden. $\endgroup$ – Benedict W. J. Irwin Nov 16 '17 at 10:02

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