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Assume the function $f(z)$ has a simple pole at $z_{0}$.

There is a theorem that states that if $C_{r}$ is an arc of the circle $|z-z_{0}| = r$ of angle $\alpha$, then $$\lim_{r \to 0} \int_{C_{r}} f(z) \, dz = i \alpha \, \text{Res}[f,z_{0}].$$

But what if $z_{0}$ is a pole of higher order? Can we say anything definitive about $ \lim_{r \to 0} \int_{C_{r}} f(z) \, dz $?

Random Variable
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2 Answers2

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For $n \geqslant 2$, the function $\dfrac{1}{(z-z_0)^n}$ has a primitive $\dfrac{(-1)}{(n-1)(z-z_0)^{n-1}}$, so if the arc is $z_0 + re^{it}$ for $\varphi \leqslant t \leqslant \vartheta$,

$$\int_{C_r} \frac{dz}{(z-z_0)^n} = \frac{1}{(n-1)r^{n-1}}\left(e^{-i(n-1)\varphi} - e^{-i(n-1)\vartheta}\right).$$

In general, that is unbounded for $r \to 0$, but for a given $n$, there are choices of $\varphi$ and $\vartheta$ that make the integral vanish.

If the principal part of $f$ has more than one term of order $< -1$, the choices for the difference between the two angles that make the integral vanish are even more restricted.

Daniel Fischer
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  • I used to think that the limit would always blow up in that situation. But then $ \displaystyle \text{PV} \int_{-\infty}^{\infty} \frac{1}{x^{3}} \ dx $ wouldn't exist. But it does exist. – Random Variable Jul 10 '13 at 18:15
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    Yes, $(n-1)(\varphi-\vartheta)$ must be a multiple of $2\pi$, then the integral along the arc vanishes. – Daniel Fischer Jul 10 '13 at 18:16
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    So the only two possibilities are that it diverges to infinity or it goes to zero? – Random Variable Jul 10 '13 at 18:19
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    Yes, you have a sum $\sum \frac{c_k}{r^k}$ for the principal part minus the $z^{-1}$ term. That is either $0$ for all $r$ (when all $c_k = 0$), or diverges to infinity dominated by the largest $k$ with $c_k \neq 0$. – Daniel Fischer Jul 10 '13 at 18:25
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I think you are going to have problems with divergences. In your example above with the double pole, you may parametrize the integral as $z=z_0+\epsilon\, e^{i \phi}$, for $\phi \in [\phi_0,\phi_0+\alpha]$. Then the integral over the arc is

$$i \epsilon \int_{\phi_0}^{\phi_0+\alpha} d\phi \, e^{i \phi}\, f(z_0+\epsilon\, e^{i \phi}) \sim \frac{i}{\epsilon} \int_{\phi_0}^{\phi_0+\alpha} d\phi \, e^{-i \phi}$$

Note that, unlike the case where you integrate above a closed curve, the integral does not vanish in general. When the integral does not vanish, the integral about a double pole (or greater, say, $n$) blows up as $\epsilon \to 0$ unless you go around a fully closed curve. (There are cases, of course, where the integral does vanish, i.e., when $n \alpha = k 2 \pi$).

Ron Gordon
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  • What about $ \displaystyle \lim_{r \to 0} \int_{C_{r}} \frac{1}{z^{3}} \ dz$ where $C_{r}$ is the upper half of a small circle about the origin? – Random Variable Jul 10 '13 at 18:28
  • Do it out. $z=r , e^{i \phi}$ and your integral becomes $$\frac{i r}{r^3} \int_{\pi}^0 d\phi , \frac{e^{i \phi}}{e^{i 3 \phi}} = \frac{i}{r^2} \int_{\pi}^0 d\phi , e^{-i 2 \phi}$$ In this case, the integral does vanish. In general, however, it does not. See my edit for clarification. – Ron Gordon Jul 10 '13 at 18:33
  • @RandomVariable: No, I just hadn't made myself clear enough. I think the post should be crystal clear now. You just have to be careful about the arcs you choose for multi-poles. The geometry usually takes care of this anyway. – Ron Gordon Jul 10 '13 at 18:48