# Contour Integration where Contour contains singularity

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?

#### Solutions Collecting From Web of "Contour Integration where Contour contains singularity"

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.