Friday, February 22, 2013

Complex Analysis: #19 Integrating Out From a Pole

  • 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

Complex Analysis: #19 Integrating Out From a Pole equation pic 1

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 ∫1f(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

Complex Analysis: #19 Integrating Out From a Pole equation pic 2

for all 0 < λ < 1.

Proof
As before, the function zλR(z) has at most finitely many poles in ℂ. Let γr, φ, where 0 < r < 1 and 0 < φ < π, be the following closed curve. It starts at the point reφi and follows a straight line out to the point Teφi, 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 γr, 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

Complex Analysis: #19 Integrating Out From a Pole equation pic 3

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