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:


About mlbaker

just another guy trying to make the diagrams commute.
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