TITLE: What is the integral of 1/(z-i) over the unit circle?
QUESTION [6 upvotes]: At present there is a simple pole on the closed contour, so the Residue Theorem appears to be inapplicable.  
But I want to claim that we can enlarge this circle to make sure that it encloses the pole, and the integral value should not change, primarily because of Cauchy's Theorem.
So the integral is simply $2\pi i$. (The residue at $z=i$ is 1.)
What do you think?  
Thanks,

REPLY [4 votes]: The integral does not converge. To show this in the most informative way, you should recall the definition of a contour integral: Let $\gamma(t)$ be a contour defined over $\mathbb{C}$ with $t\in I$. Then
\begin{align}
\oint_{\gamma}{f(z)dz}:=\int_I{f(\gamma(t))\gamma'(t)dt}.
\end{align}
So applying this to the integral in question (I've missed out a bit of algebra), we get
\begin{align}
\oint_{S^1}\frac{1}{z-\imath}dz&=\int_0^{2\pi}{\frac{\imath}{e^{\imath t}-\imath}}dt\\
&=\int_0^{2\pi}\frac{\sin(t)-1+\imath\cos(t)}{1-2\sin(t)}dt.
\end{align}
Now, one can (after a little work) compute that
\begin{align}
\int\frac{\sin(t)-1+\imath\cos(t)}{1-2\sin(t)}dt=\frac{\tanh^{-1}\left(\frac{\tan\left(\frac{t}{2}\right)-2}{\sqrt{3}}\right)}{\sqrt{3}}-\frac{t}{2}-\frac{\imath}{2}\log(1-2\sin(t))+z_0.
\end{align}
This function is undefined at both $0$ and $2\pi$ and hence the integral has no limit, or formally it diverges.