# Calculation of a certain type of contour integral

1. Jul 15, 2010

### hunt_mat

Hi,

During my research I came across a contour integral where the pole was on the boundary. I have never come across this before, do anyone of you know how I would go about computing this?

It involved the Hilbert transform and I can't find it in my undergraduate complex analysis books and I thought someone here might know.
On a similar note, if anyone has any good numerical routines for Hilbert transform, I would like to know.

Mat

2. Jul 15, 2010

### Mute

The way such a thing is dealt with is to have the contour run around the pole in a small semi-circular arc, in the limit that the radius of the arc tends to zero.

As long as it's a simple pole, the result is if a contour runs straight through a pole (which would correspond to a half-circle arc that excludes the pole from being inside the contour) the pole contributes half the residue to the integral. If the pole is at a corner of a contour (which corresponds to a quarter-circle arc around the contour) it would contribute 1/4 the residue.

3. Jul 16, 2010

### hunt_mat

Do you have a reference? The same idea is used in deriving the solution of the Benjamin-Ono equation.

Surely if the pole was outside of the contour the integral would be zero by Cauchy's theorem. Am I missing something?

4. Jul 16, 2010

### jackmell

5. Jul 16, 2010

### jackmell

6. Jul 16, 2010

### Mute

No, it's not zero because in the limit that the arc around the pole goes to zero the contour goes straight through the pole, so there is a contribution.

Consider:

$$\oint_c dz \frac{f(z)}{z-z_0}$$
where f(z) is analytic everywhere inside and on the contour. If the pole of the integrand at z_0 lies on our contour C, we go around it in a semi-circular arc of radius $\epsilon$, as $\epsilon \rightarrow 0$. This part of the curve we'll parametrize by $z-z_0 = \epsilon \exp(i\theta)$. This gives

$$\int_\pi^0 d\theta i\epsilon e^{i\theta} \frac{f(z_0 + \epsilon e^{i\theta})}{\epsilon e^{i\theta}}$$

If we now take the limit $\epsilon \rightarrow 0$, we get
$$-i\int_0^\pi d\theta f(z_0) = -i\pi f(z_0)$$
which is half of the residue of f(z) at z_0 (with a minus sign because we traversed over the pole in a clockwise rotation).

Note that this calculation wouldn't work if the pole weren't simple, i.e., if the pole were $(z-z_0)^m$, with $m > 1$.

(However, jackmell's link suggests that if the pole is an odd power and our contour through it is a straight line segment, then treating the straight line segment as a principal value integral gives the half-residue again.)

That is interesting. I had not heard of this result before. Also interesting is that the contour through the pole in this case must be a straight line segment.

Last edited: Jul 16, 2010