Mathematics of Physics and Engineering

Residue Integration 229

As a result,

lim IR =

R—+00

EXERCISE 4A.11F (a) Without using a computer, verify that

dx wy/3

I _ 00 (4 + 2x + x^2 )^3 72 '

(b) Convince yourself that a computation of (4-4-18) using a mirror image

of the curve CR in the lower half-plane leads to the same final answer. Hint:

the reason is that the roots of Q come in complex conjugate pairs.

The same method applies to the computation of integrals

f°° P(x) f°° P(x)
/ cos(a:r) dx and / sinfaa;) , dx, (4.4.19)
7-oo Q{x) 7-oo Q(x)
where a is a real number, P, Q are polynomials without common roots, the
degree of Q is at least two units bigger than the degree of P, and Q has no
real roots. Assuming that a > 0, you evaluate the integral

/

OO

Aax

W)

dx (4A20)

using the steps described above, and then take the real or imaginary part
of the result. The integral in (4.4.20) can exist even when the degree of Q
is only one unit bigger than the degree of P, see Problem 5.8 on page 436.
EXERCISE 4.4.12.c (a) Evaluation of the integrals in (4-4-19) will not go
through if, instead of eiaz you try to work directly with cos az or sin ax.
What goes wrong? Hint: look carefully at cosaz and sin az when \z\ = R. (b)
How should you change the integral in (4-4-20) if you want to evaluate
(4-4-19) with a > 0 by integrating over a semi-circle with Sz < 0? Hint:
you should go with e~iax. (c) Without using a computer, verify that

f

J — t

COS X IT

1+x^2 dx = —.

Then ask your computer algebra system to evaluate this integral.

Some real integrals are evaluated using residues and integration over the