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I'm trying to clarify some thoughts on contour integration.

If I have an integral $\int_{c-i\infty}^{c+i\infty} f(z) dz$, where $f(z)$ has finitely many poles in the complex plane...is this equivalent to doing the integral on a contour that encircles the singularities counter-clockwise? (i.e. is the answer to the above integral given by the residue theorem?)

Furthermore, if there were poles at all of the negative integers for $f(z)$, would the difference between the above integral and adding the residues at finitely many poles, be the sum of the excluded residues?

I think the answer is yes. I know when one is doing a real integral, a typical method is to create a half-circle with the diameter on the x-axis, and allow the limit of the radius to go to infinity, while showing the contribution from circular portion tends to 0. Will the same kind of argument work in the above case in general?

I also think the answer is yes based on how I understand the Riemann Sphere interpretation of the complex plane. Any "infinite" contour is really a loop in the complex plane anyway because it's "passing through the point at infinity."

Thanks for your help.

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  • $\begingroup$ Short answer: That depends on the behaviour of $f$ at infinity. Try the substitution $z=1/\zeta$ to see that in more detail. $\endgroup$ – Harald Hanche-Olsen Nov 24 '14 at 22:08
  • $\begingroup$ Thanks. I think that something with an essential singularity at infinity would make my question impossible to answer in general. $e^z$ for instance as $z \rightarrow \infty$ could either be 0 or infinity. Is that right? So then, do I need to restrict my questions to cases where it is finite at infinity or has a pole at infinity? $\endgroup$ – MathStudent Nov 24 '14 at 22:13
  • $\begingroup$ If I did that substitution, it seems my limits of integration would just be 0? $\endgroup$ – MathStudent Nov 24 '14 at 22:18
  • $\begingroup$ Well, it's a closed path after all, so that makes sense. You would have to pay attention to that path, however. And the possible resulting singularity, which used to be at $\infty$ but is now at $0$ instead. $\endgroup$ – Harald Hanche-Olsen Nov 25 '14 at 7:57
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if you set: $$ w = \frac{z-c}{i} $$ then the integral takes the more familiar form: $$ I = i\int_{-\infty}^{\infty} g(w)dw \\ \text{where } g(w)=f(iw+c) $$ which shows that the usual arguments apply (having appropriately relocated the poles).

behaviour at infinity

if you want to use the residue theorem to evaluate (as a limit) the integral along the real axis (corresponding to the original line $(c-i\infty,c+i \infty)$ the requirement is that the integral around the rest of a complete circuit should tend to zero in the limiting case. this is guaranteed, for example in the simple case where the integrand is a rational function for which the degree of the denominator is two greater than the degree of the numerator.

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