## Behaviour of residues

So the other day, I was talking to a friend of mine about complex analysis — particularly, the application of the residue theorem to the evaluation of definite (real) integrals. He stated the following (I’ve paraphrased/tried to make it precise):

Theorem 1. Provided that $zf(z) \to 0$ as $|z| \to \infty$ and $f(z)$ is holomorphic on all of $\mathbb{C}$ except possibly with finitely many isolated singularities occurring in $\mathbb{C} \setminus \mathbb{R}$, we have that

$\displaystyle \int_{-\infty}^\infty f(x) \, dx = 2\pi i \sum_a \mathrm{Res}(f,a)$,

where the sum is taken over all $a$ in the upper half-plane.

The condition that $zf(z) \to 0$ as $|z| \to \infty$ simply ensures that we can take the semicircle $\alpha$ which starts at $-r$, goes to $r$, and then arcs back to $-r$ in a counterclockwise direction, and letting this $r$ go to infinity, the contour integral of $f(z)$ around the curve $\alpha$ will indeed converge to the desired improper integral (the left-hand side of the above equation). That is, it ensures that the contribution of the “arc” portion of the curve to the integral converges to zero as $r \to \infty$.

After he said this, I started thinking about the following theorem:

Theorem 2. If a holomorphic function has finitely many singularities, then the sum of all the residues must be 0.

This means that if a function has no singularities on the real line, then the sum of its residues in the upper half-plane must be the negation of the sum of its residues in the lower half-plane.

Going back to the discussion of Theorem 1 now. We could also take the semicircle in the bottom half-plane, whereby the curve would wind around any enclosed singularities in a clockwise direction. The residue theorem will then spit out a negative sign in front of the sum, since the winding number will be $-1$ (due to the clockwise rotation):

$\displaystyle \frac{1}{2\pi i} \int_{-\infty}^\infty f(x) \, dx = \sum_{\substack{\text{upper} \\ \text{halfplane}}} \mathrm{Res}(f,a) = -\sum_{\substack{\text{lower} \\ \text{halfplane}}} \mathrm{Res}(f,a)$

whereby we observe the residues in the upper and lower half-planes summing to zero as expected.

A quick corollary of this is the following: if $p(z)$ is a polynomial with $\mathrm{deg} \; p(z) \geq 2$ with all of its roots in one particular half-plane, then $f(z) = 1/p(z)$ clearly satisfies the limit condition, but then the sum on the right-hand side of the equation is zero by Theorem 2, which means that

$\displaystyle \int_{-\infty}^\infty \frac{dz}{p(z)} = 0.$

I’ll write another post soon; right now I have to sleep. Here’s a plot of
$\displaystyle \left| \int_{-t}^t \frac{1}{(x-(3+i))(x-(2+9i))} \, dt \right|$ for $0 \leq t \leq 200$: