# Contour Integral

1. Nov 15, 2014

### eyesontheball1

Is it possible to evaluate the integral of log(z) taken over any simple closed contour encircling the origin? I don't fully understand how singularities on branch cuts should be treated when integrating over contours encircling such singularities. Are residues applied? Can someone explain this to me? Thanks!

2. Nov 16, 2014

### HallsofIvy

Looking at your specific example, ln(z), by convention we take the branch cut along the negative real axis or $\theta= \pi$ when [itex]z= re^{i\theta}[/tex]. And, of course, $$ln(z)= ln(r)+ i\theta$$. So integrating across a branch cut results in adding [tex]2\pi n i[tex] for some integer n. That is the difficulty with integrating over branch cuts- the integral value jumps by some multiple of a constant.

3. Jan 18, 2015

### Svein

If you put z = r*(cos(φ) + i*sin(φ)), log(z) = ln(r) + i*φ (since both sine and cosine are periodic with period 2π, there are several values of φ we can use in the expression).

Thus ∫log(z) for |z|=R, is equivalent to ∫(ln(R) +i*φ)dφ where the integration limits are 0 and 2π. The rest is left as an exercise for the student...