# Residue at $z=\infty$

I'm a bit confused at when to use the calculation of a residue at $z=\infty$ to calculate an integral of a function.

Here is the example my book uses: In the positively oriented circle $|z-2|=1$, the integral of $$\frac{5z-2}{z(z-1)}$$ yields two residues, which give a value of $10\pi i$ for the integral, using the Cauchy Residue Theorem. I've got that down.

The book later, calculates the residue at infinity, yielding the same answer... - one residue calculation!

I can't seem to find what I'm missing here... Why is it that when considering this one bound, $2 \lt |z-2|\lt\infty$, we can use the residue at infinity to find the value of the integral... yet at the same time calculate a residue in three different bounds using the Cauchy Integral Theorem and find the same integral value?

Thanks!

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The integral can be computed by the sum of the residues outside of the curve; in this case, there is only the one at infinity. As you can see, the sum of all the residues is $0$ (for any function with finitely many singularities on the Riemann sphere). There are many curious identities you can derive in this way. – user8268 May 15 '12 at 6:03
Did you mean $|z-1|=2$? The circle you wrote down doesn't contain the pole at $z=0$ and goes through the pole at $z=1$. – joriki May 15 '12 at 7:10
to extend the excellent comment by @user8268: you can perform the variable substitution $u=z^{-1}$ which interchanges the outside and the inside of the contour. Then you can calculate the "usual" residue at the origin. – Fabian May 15 '12 at 11:29
Joriki, actually it's |z|=2! I grabbed the other circle from a nearby example... Thx! – user5262 May 15 '12 at 16:56
@user8268: That's interesting... is there a reason why, in a nutshell? – user5262 May 15 '12 at 16:58

When can you use the residue at $\infty$ to calculate the value of an integral $\int f$ ?

is basically given by the following result.

If $f$ is a holomorphic function in $\mathbb{C}$, except for isolated singularities at $a_1, a_2, \dots , a_n$, then $$\operatorname{Res}{(f; \infty)} = -\sum_{k = 1}^{n} \operatorname{Res}{(f; a_k)}$$

were the residue at infinity is defined as in the wikipedia article. Now, from Cauchy's residue theorem it follows that

$$\operatorname{Res}{(f; \infty)} = -\frac{1}{2 \pi i} \int_{\gamma} f(z) \, dz$$

where you can take $\gamma$ to be a circle $|z| = R$, where $R$ is large enough so that all the singularities $a_k$ are contained inside the circle. Of course you can then use the more sophisticated versions of Cauchy's integral theorem to change the curve $\gamma$, but this one suffices for simplicity.

Then the last formula gives you a way to calculate a complex integral just by calculating the residue at infinity of the function, instead of computing all the "finite" residues.

Now to answer your second question of why this is the case, maybe a sketch of a proof of this result will be enough.

So let $\displaystyle{F(z) := -z^{-2}f(z^{-1})}$, then since $f(z)$ is holomorphic for $|z| > R$ for some large enough $R$, we see that $F(z)$ is holomorphic for $|z^{-1}| > R$, or equivalently, for $0 < |z| < \frac{1}{R}$. Thus $F$ has an isolated singularity at the origin, and then by the definition of the residue at infinity we have

$$\operatorname{Res}{(f; \infty)} := \operatorname{Res}{(F; 0)} = \frac{1}{2 \pi i} \int_{|w| = \frac{1}{R}} F(w) \, dw = \frac{1}{2 \pi i} \int_{|w| = \frac{1}{R}} -\frac{f(w^{-1})}{w^2} \, dw$$

Then by making the substitution $z = \frac{1}{w}$ we get

$$\frac{1}{2 \pi i} \int_{|w| = \frac{1}{R}} -\frac{f(w^{-1})}{w^2} \, dw = \mathbf{\color{red}{-}} \frac{1}{2 \pi i} \int_{|z| = R} \, f(z) dz$$

where the last negative sign comes from the fact that the new circle $|z| = R$ you get after the substitution has its orientation reversed. And this last equality is precisely what we wanted to prove.

Note

This result is exercise 12 in section V.2 in Conway's book Functions of One Complex Variable I (page 122), or it also appears in exercise 6 in section 13.1 of Reinhold Remmert's book Theory of Complex Functions (page 387), in case you want some references.

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Thank you! I like that I can avoid using the Cauchy residue theorem. What if at least one singularity lies ON the given circle, can I still use this theorem? Yes, yes, I think I can. Right? So, for an integral of (z^5)/((1-(z^3)) it would be more prudent to find the value using the residue at z=0 in the integrand of (1/(z^2))f(1/z) - where C is positively oriented circle |z|=2? – user5262 May 16 '12 at 4:15