• Complex Analysis: #19 Integrating Out From a Pole

Maybe we are dissatisfied with this “Cauchy principle value” technique. After all, it is rather like cheating! So let’s see what we can do with an integral like

where 0 is a pole, and ∞ is a zero of f. For example, look at the function f(x) = 1/x. But here we see big problems! Both of the integrals ∫₁^∞ f(x)dx and  ∫₀¹ f(x)dx are divergent.

Thinking about this, we see that the problems with the function 1/x stem from the fact that, first of all, the zero at ∞ is simple, and second of all, the pole at 0 is simple. This leads us to formulate the following theorem.


Theorem 39
Again, let R be a rational function defined throughout ℂ, but this time with a zero of order at least 2 at ∞. Furthermore, R has no poles in the positive real numbers (x > 0), and at most a simple pole at 0. Then we have

for all 0 < λ < 1.

Proof
As before, the function z^λR(z) has at most finitely many poles in ℂ. Let γ_{r, φ}, where 0 < r < 1 and 0 < φ < π, be the following closed curve. It starts at the point re^φi and follows a straight line out to the point Te^φi, where T = 1/r. Next it travels counter-clockwise around the circle of radius T, centered at 0, till it reaches the point Te^(2π−φ)i. Next it travels along a straight line to the point re^(2π−φ)i. Finally it travels back to the starting point, following the circle of radius r in a clockwise direction. Lets call these segments of the path L₁, L₂, L₃ and L₄. By choosing r and φ to be sufficiently small, we ensure that the path γ_{r, φ} encloses all poles of the function. In the exercises, we have seen that the path integrals along the segments L₂ and L₄ tend to zero for r → 0. Thus we have