# Complex Analysis: #19 Integrating Out From a Pole

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 ∫

_{0}

^{1 }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

^{λ}R(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 L

_{1}

_{}, L

_{2}, L

_{3}and L

_{4}. 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 L

_{2}and L

_{4}tend to zero for r → 0. Thus we have

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