Mathematics of Physics and Engineering

230 Singularities of Complex Functions

curve CR^ in Figure 4.4.1(b). FOR EXAMPLE, let us evaluate the integral

cos (3a;) — cos a;

Jo

dx. (4.4.21)

Step 0. EXERCISE 4.4.13. c Verify that the integral converges. Hint:
f?° dx/x^2 for every 5 > 0, and, for x near 0, cos(ax) = 1 — a x /2 + ....
Step 1. By the Integral Theorem of Cauchy, we have

JCR,P Z

Step 2. The integrals over the parts of CR,P on the real axis result in

f — p pSix cix pR -Zix „ix

/

—p e3iz _ eix r

dx

P x*

rR / „3ix i „-3tx pix I p-ix\

= / f^—£r J J <fa - 2f, P -> o, it -> oo.

Step 3. The integral over the semi-circle of radius R tends to zero as

R —> oo. All you need to notice is that, for z — R(cos8 + ism8), we have

\piaz\ __ p — aRs'm& < i

Step 4- The integral over the semi-circle of radius p tends to — 2n as

p —> 0. Indeed, the Taylor expansion shows that, near z = 0, we have

z~^2 {e3lz —elz) = 2iz~l +g(z), where g(z) is analytic at z = 0. The integral

of g will then tend to zero, and the integral of 2iz-1, once you take into

account only half of the circle, the clock-wise orientation and the factors of

i, produces 27r.

Step 5. By combining the results of the above steps, we conclude that

f

Jo

cos(3a;) — cos x ,

T ax = —n.

EXERCISE 4.4.14. (a)A Provide the details in the steps 3-5 above. (b)G Is

it possible to evaluate (4-4-&1) by integrating (cos(3;z) — cosz)/,?^2 over the

curve in Figure 4-4-l(a)?

With all these new techniques of integration, we should never forget the

basic rules related to symmetry. First and foremost, the integral of an odd

function (that is, a function / = f(x) satisfying f(x) = —f(—x)) over a