integral of $1 \over |z|$ over a circle

Question

I am terribly confused by complex integrals involving $|z|$. Please help me. Let's say I want to evaluate $$ \int_{a<|x|<b} {dx \over |x|} \quad x \in \mathbb R $$ by contour integration. I choose to close the path by adding the arc of radius $b$ (counter-clock wise), call it $\Gamma$, and the arc of radius $a$ (clock-wise), call it $\gamma$, on the upper half-plane. So by Cauchy theorem, the integral on this closed contour is 0, since there is no singularity in it.

But on $\Gamma$, we have $$ \int_\Gamma {dz \over |z|} = \int_0^\pi {b i e^{i \theta} \over b}d\theta = -2 $$ Similarly $$ \int_\gamma {dz \over |z|} = 2 $$ Thus I conclude $$ \left( \int_{-b}^{-a} + \int_{a}^{b} \right) {dz \over |z|} = 0 $$ And of course I know this is wrong -- the integral is 2 $\ln(b/a)$. What am I doing wrong? Thank you for your help!

Answer 1

Trying to make some sense of your question: if $\;0<a<x<b\;$ , then we simply have

$$\left.\int_a^b\frac{dx}x=\log x\right|_a^b=\log \frac ba$$

If $\;b<0\;$ it is something similar. If $\;a\le 0<b\;$ , then you have a divergent improper integral....

Answer 2

Now I see what I am doing wrong. $1/|z|$ is not holomorphic because $$ {1 \over |z|} = {1 \over \sqrt{x^2 + y^2}} + 0 \times i = u(x,y) + i v(x,y) $$ Since $$ {\partial u(x,y) \over \partial x} \neq {\partial v(x,y) \over \partial y} $$ the Cauchy-Riemann equations are not satisfied and the Cauchy theorem is not applicable. Thank you all for your help!