# Why is $\int\limits_{\gamma} \frac{1}{z-1} \neq 2\pi i$, $\gamma = \{z : \lvert z \rvert = 1\}$?

$\int\limits_{\gamma} \frac{1}{z-1}$

$\gamma = \{z : \lvert z \rvert = 1\}$

I use Cauchy's integral formula, which says $$\int\limits_{\gamma} \frac{f(z)}{(z-a)^{n+1}} = \frac{2\pi i}{n!} f^{(n)}(a)$$ and I choose $f(z) = 1$ to be my holomorphic function.

$n = 0$ and $a = 1$ (which doesn't matter since my $f(z)$ is a constant function), so evaluating Cauchy's integral formula, $$\int\limits_{\gamma} \frac{1}{z-1} = 2\pi i.$$

Why is this not the correct solution?

• The path $\gamma$ in Cauchy's formula should surround $a$. Not pass through it as in your case. – GEdgar Apr 17 '15 at 1:42
• There is a rather surly singularity at $z=1$; your curve passes right through that. All bets are off. – ncmathsadist Apr 17 '15 at 1:44

If there is a pole on the contour, the integral doesn't converge. If it is a simple pole, you can find the Cauchy principal value (basically, if the integrand has a singularity at $c \in [a,b]$, the CPV is $$\lim_{\varepsilon \to 0} \int_a^{c-\varepsilon}+\int_{c+\varepsilon}^b)$$ by taking half the residue at $c$, which you prove by deforming the contour, adding a small semicircle around $c$.
• A simple pole is just one where the most singular term is $(z-a)^{-1}$, so yes. – Chappers Apr 17 '15 at 2:31