# Contour integration when pole is outside the contour

Here they are using the pole OUTSIDE the contour? I thought this was illegal according to the residue theorem or we are not supposed to do contour integration with poles outside the contour itself.

• Which pole do you mean? They are not using the pole at zero, they are noticing that there are relevant poles because of the fact that the pole at zero isn't inside the contour. – Ian Dec 23 '14 at 19:51
• So they're not using the residue theorem? – Amad27 Dec 23 '14 at 20:01
• Well, they do: the residue theorem says the integral over the contour is $0$. – Daniel Fischer Dec 23 '14 at 20:04
• @DanielFischer, how? Where?? – Amad27 Dec 23 '14 at 20:07
• There is no use of the Residue theorem here. It's Cauchy's integral Theorem that the integral is zero. This is because the integrand is analytic inside of the closed contour because there are no singularities there. Next, they parameterize the contour accordingly. The rest of the problem shouldn't be too bad. – DaveNine Dec 23 '14 at 21:45

In this case you can use the Sokhotski–Plemelj theorem and work with the principal value directly. This amounts to putting the pole right on the contour and counting it as being "half inside" making the contribution $\pi i$ times the residue. In general, a contour integral can be written as minus $2\pi i$ times the sum of all the residues of the poles outside the contour, but you then also need to include the residue of the pole at infinity (if there is a pole there). To see this consider an arbitrary contour integral of a meromorphic function $f(z)$:

$$I = \oint_C f(z) dz$$

where $C$ is a counterclockwise contour. Suppose that a point $p$ is inside the contour. We can then define a new variable $w$ by putting:

$$w = \frac{1}{z - p}$$

Performing this change of variables to the integral yields:

$$I = \oint_{C_w} \frac{1}{w^2}f\left(\frac{1}{w}+p\right) dw$$

Note that the transform maps the interior of the contour to the exterior and vice versa, the winding number of the contour reverses, but we absorbed a minus sign from the transform of $d$z to $dw$ to keep the new contour counterclockwise.

Suppose then that $f(z)$ has a pole of order m at $z = q$. This will then correspond to a pole at $w = q_w = \frac{1}{q-p}$ of $f\left(\frac{1}{w}+p\right)$. The Laurent expansion of $f(z)$ transforms to the Laurent expansion for $\frac{1}{w^2}f\left(\frac{1}{w}+p\right)$ as follows. A term $(z-q)^{-n}$ transforms to the term $(-1)^n q_w^n w^{n-2}(w-q_w)^{-n}$. One then has to expand the factor $w^{n-2}$ around $q_w$ to obtain the transformed Laurent expansion. But note that for $n>1$ one obtains a polynomial of degree $n-2$ multiplying the term $(w-q_w)^{-n}$, this then yields contribution up to the term $(w-q_w)^{-2}$ but not to higher powers, so the residue is not affected by such terms. The residue is thus obtained from only the term $(-1)^n q_w^n w^{n-2}(w-q_w)^{-n}$ for $n = 1$ expanded around $q_w$, this yields the term $-\frac{1}{w-q_w}$. This means that the residue of the new integrand changes sign relative to original residue of $f(z)$.

We know that the integral is the sum of $2\pi i$ times the sum of the residues of the poles inside the contour, if we apply this to the transformed integral we see that this is the same as minus $2\pi i$ times the sum of the residues of the poles outside the original contour of the original integral. But this is only true if there is no pole at $w = 0$. If there is a pole there then we define the residue at that point as minus the residue of $f(z)$ at infinity. This then makes the statement that the integral is minus the sum of the residues outside the contour true in general.