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Why, when calculating contour integrals, we need to avoid the singularities inside the region delimited by the closed path we're integrating over? Wouldn't we need to care only about singularities laying on the actual closed path?

Joan S. Guillamet F.
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  • if a singularity is on the integration path, your function would be infinite at this point... – Jean Marie Dec 01 '23 at 22:22
  • Right, that's what I said: why not care only for whether there are singularities on the closed path and not inside the region delimited by it? – Joan S. Guillamet F. Dec 01 '23 at 22:26
  • For instance, $$\oint_{|z|=R}\dfrac{1}{z}dz\ (R\neq 0),$$ Why even care about the singularities on $|z|\leq R$ and not only $|z|=R$ (on the actual closed path)? – Joan S. Guillamet F. Dec 01 '23 at 22:29
  • Only points around which you make a complete turn $\color{red}{2 \pi}$ will leave a "memory" of $\color{red}{2 \pi}ir$ (where $r$ is the residue). I strongly advise you to read "Visual complex analysis" (T. Needham) where this sort of things is described. One could consider, in some cases, that a point on the contour corresponds to a turn of $\color{red}{\pi}$ only. – Jean Marie Dec 01 '23 at 22:33
  • In the example you just gave, there is no singularity of function $1/z$ on the contour... – Jean Marie Dec 01 '23 at 22:37
  • I know it doesn't (sorry for the poor example), it was only to know why we have to avoid, in this case, $z=0$ if it isn't on the closed path. – Joan S. Guillamet F. Dec 01 '23 at 22:59
  • We don't have to avoid $z=0$ if it isn't on the path. Jean Marie tried to highlight to you what you should be looking for. The amazing thing about this integral is that it tells us something about how many times the path goes around the origin. "winding number" is another important buzz word. Also have you thought about the integrals $$\oint_{|z|=r}\frac1{z^n},dz,,n>1,?$$ Why is $n=1$ so special? – Kurt G. Dec 02 '23 at 07:34
  • If you want to know why we are so interested in how many times paths go around singularities you may want to watch the very good video OP has linked in this post. – Kurt G. Dec 02 '23 at 07:39
  • Well, the antiderivative of $\dfrac{1}{z}$ is $\log z$, which is multivalued, hence the path passes through a branch cut – Joan S. Guillamet F. Dec 02 '23 at 09:42
  • Exactly. And for $n>1$ the antiderivative is $\frac{-1}{(n-1)z^{n-1}}$ which does not have these problems. This has a lot of analogies with conservative vector fields, closed/exact differential forms and the index theory that is built on those. – Kurt G. Dec 02 '23 at 10:07
  • The reason that they are redrawing the contour so that there are no singularities inside is that they are preparing to apply Cauchy's Theorem, which says that if there are no singularities inside the closed contour, the integral will be $0$. From this, you can deduce that when there are singularities, the contour integral is equal to the sum of integrals along circles about each of the singularities. And as those circles can be any size, it is very easy to figure out what their values have to be without actually calculating that integration. This proves the very powerful Residue Theorem. – Paul Sinclair Dec 02 '23 at 23:21

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