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Complex Analysis - Maths KU

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Important observation about even<br />

functions<br />

C R<br />

z 3<br />

z 2<br />

–R 0<br />

y<br />

z n<br />

z 1<br />

z 4<br />

x<br />

R<br />

Figure 6.11 Semicircular contour<br />

☞<br />

6.6 Some Consequences of the Residue Theorem 355<br />

The limit in (9), if it exists, is called the Cauchy principal value (P.V.) of<br />

the integral and is written<br />

� ∞<br />

� R<br />

P.V. f(x) dx = lim<br />

−∞<br />

R→∞<br />

f(x) dx.<br />

−R<br />

(11)<br />

In (10) we have shown that P.V. � ∞<br />

xdx= 0.To summarize:<br />

−∞<br />

Cauchy Principal Value<br />

When an integral ofform (2) converges, its Cauchy principal value is the<br />

same as the value ofthe integral. Ifthe integral diverges, it may still<br />

possess a Cauchy principal value (11).<br />

One final point about the Cauchy principal value: Suppose f(x) is continuous<br />

on (−∞, ∞) and is an even function, that is, f(−x) =f(x).Then<br />

its graph is symmetric with respect to the y-axis and as a consequence<br />

� 0<br />

� R<br />

f(x) dx = f(x) dx (12)<br />

and<br />

−R<br />

� R<br />

−R<br />

0<br />

� 0<br />

f(x) dx = f(x) dx +<br />

−R<br />

� R<br />

0<br />

f(x) dx = 2<br />

� R<br />

0<br />

f(x) dx. (13)<br />

From (12) and (13) we conclude that if the Cauchy principal value (11) exists,<br />

then both � ∞<br />

0 f(x) dx and � ∞<br />

f(x) dx converge.The values of the integrals<br />

−∞<br />

are<br />

� ∞<br />

� ∞<br />

� ∞<br />

f(x) dx =P.V. f(x) dx.<br />

0<br />

f(x) dx = 1<br />

2 P.V.<br />

� ∞<br />

f(x) dx and<br />

−∞<br />

To evaluate an integral � ∞<br />

f(x) dx, where the rational function f(x) =<br />

−∞<br />

p(x)/q(x) is continuous on (−∞, ∞), by residue theory we replace x by the<br />

complex variable z and integrate the complex function f over a closed contour<br />

C that consists of the interval [−R, R] on the real axis and a semicircle CR of<br />

radius large enough to enclose all the poles of f(z) =p(z)/q(z) in the upper<br />

half-plane Im(z) > 0. See Figure 6.11. By Theorem 6.16 of Section 6.5 we<br />

have<br />

� �<br />

� R<br />

n�<br />

f(z) dz = f(z) dz + f(x) dx =2πi Res(f(z),zk),<br />

C<br />

−R<br />

CR<br />

where zk, k =1,2,...,n denotes poles in the upper half-plane.If we can<br />

show that the integral �<br />

f(z) dz → 0asR→∞, then we have<br />

CR<br />

� ∞<br />

� R<br />

P.V. f(x) dx = lim<br />

−∞<br />

R→∞ −R<br />

−∞<br />

f(x) dx =2πi<br />

k=1<br />

−∞<br />

n�<br />

Res(f(z),zk). (14)<br />

k=1

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