There are many theorems in complex analysis which tell us about integration $\int_{\gamma} f$ where $f$ is continuous (or even differentiable) in the interior of $\gamma$ except finitely many points.
I would like to see how should we solve $\int_{\gamma}f(z)dz$ where $\gamma$ contains a singular point of $f$.
For example, in solving $\int_{|z|=1} \frac{1}{z-1}dz$, I slightly pulled the curve $|z|=1$ near $1$ towards left, so that $\gamma$ will not contain $1$; then the singularity of $\frac{1}{z-1}$ will not be inside this deformed curve, so integration over deformed curve is zero, so taking limit, I concluded that $\int_{|z|=1} \frac{1}{z-1}dz=0$.
However, if I pull the curve near $1$ on the right side, then the singularity of $\frac{1}{z-1}$ will be inside this deformed curve, and integration is then non-zero.
I confused between these two processes; can you help me what is the correct way to proceed for finding integration along curves, in which singularity of the function is on the curve?
Best Answer
As an improper integral, $\oint_{|z|=1} \frac{1}{z-1}\; dz$ does not converge. You could define a principal value which is the limit as $\epsilon \to 0+$ of the integral over arcs omitting a length $\epsilon$ on each side of the singularity. But you have to be careful doing such an integral using residues, because the arc you use to close up the contour will make a significant contribution.