Find $\oint_C \frac{dz}{z-2}$ on the square $C$ with vertices $\pm2\pm 2i$ .
As there is a pole at $z=2$, I removed it by taking a semicircle of small radius $r$ about $2$ and the integral on the resulting curve should be zero. The integeral on the semicircle should be $\pi i$. When $r$ tends to zero, the required integral should be $\pi i$. I am not sure if I am doing this right.
The integral along one eighth of the contour is $$\int_{2+\epsilon i}^{2+2i}\frac{dz}{z-2}=\int_\epsilon^2\frac{i\,dy}{iy}=\ln2-\ln\epsilon\to\infty\ \hbox{as $\epsilon\to0^+$}\ ,$$ so this integral, and therefore the whole integral, diverges.
If you are looking for a principal value, or if you are using some other definition for convergence of an improper integral, you need to state this clearly in your question.