# Complex Analysis: #18 Integrating across a Pole

Obviously this is nonsense, since the integrals from −1 to 0, and from 0 to 1 of the function 1/x diverge. Specifically, for 0 < ∈ < 1 we have

Thus, if we agree to abandon the principles we have learned in the analysis lectures, and simply say that

then we have the

*Cauchy principle value*of the integral, and we see that in this case it is simply zero. One could make an important-looking definition here, but let us conﬁne our attention to integrals along closed intervals [a, b] ⊂ ℝ of complex-valued functions, where there might be poles of the function in the given interval. Assume for the moment there is a single pole at the point p ∈ (a, b). Then we will define the

*principle value*of the integral (if it exists) to be

Then the generalization to having a ﬁnite number of poles of f along the interval (but not at the endpoints) is clear.

**Theorem 38**Let R be a rational function, defined throughout ℂ (together with it’s poles). Assume that it has a zero at inﬁnity, so that there can only be ﬁnitely many poles. Let p

_{1}< . . . < p

_{m}be the poles of R which happen to lie on ℝ. Assume that each of these poles is simple; that is, of order 1. We distinguish two cases:

- If R has a simple zero at inﬁnity (that is, of order 1), then we take f(z) = R(z)e
^{iz}. - Otherwise, R has a pole of order at least 2 at inﬁnity, and in this case we take f(z) = R(z).

Then we have

*Proof*

In either case, we can have only ﬁnitely many poles of the function f; therefore only ﬁnitely many poles along the real number line. Let γ

_{δ}be the path along the real number line from −∞ to ∞, but altered slightly, following a semi-circle of radius δ above each of the poles on the real line. Furthermore, δ is sufficiently small that no other pole of f is enclosed within any of the semi-circles. Then, according to our previous theorems, we have

To simplify our thoughts, let us ﬁrst consider the case that there is only one single pole p ∈ ℝ on the real number line. And to simplify our thoughts even further, assume that p = 0. Then we have the path γ

_{δ}coming from −∞ to −δ, then it follows the path δe

^{iπ(1−t)}, for t going from 0 to 1, and then ﬁnally it goes straight along the real number line from δ to ∞.

Let us now consider the Laurent series around 0. We can write

say. But then we can just define the new function g throughout ℂ (leaving out the ﬁnite set of poles of f) by the rule

Obviously g has no pole at 0, but otherwise it has the same set of poles as the original function f. Since the function c

_{−1}/z is analytic at all these other poles, the residue of g is identical with that of f around each of these poles. Therefore

Now, to get the principle value of the integral for f, we use the path γ

_{δ}, but we must remove the semi-circle part of it. That is, let

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