1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

302 CHAPTER 8 • RESIDUE THEORY

Suppose that we want to evaluate an integral of the form

{2"

lo F (cosO,sinO) dO, (8-3)

where F (u, v) is a ftmction of the two real variables u and v. Consider the unit

circle Ci (0) with parametrization

Ci (O): z = cosO + isinO = ei9, for 0 ~ (} ~ 27r,

which gives us the symbolic differentials

dz= (- sinO + icosO) dO = i ei^8 dO and

d(}= dz= dz_

ie•^8 iz

Combining z = cosO + i sin 0 with ~ = cosO - i sin 0, we obtain

cos 0 = ~ ( z + ; ) and sin 0 = ;i ( z - ; ).

(8-4)

(8-5)

Using the substitutions for cosO, sinO, and dO in Expression (8-3) transforms the

definite integral into the contour integral

fo

2

" F (cosO, sinO) dO = j f (z) dz ,

ct(o)

where the new integrand is f (z) = F(Hz+~]~H·- f))_

Suppose that f is analyt ic inside and on the unit circle C 1 (0), except at the

points zi, z2, ... , z,, that lie interior to Ci (0). Then the residue theorem gives

{2" n

lo F (cosO, sinO) dO = 27ri 2: Res [J,zk].

0 k = I

(8-6)

The situation is illustrated in Figure 8.2.

• EXAMPLE 8.10 Evaluate J~" 1+ 3 ~0<; 2 8 d0 by using complex analysis.
  Solution Using Substitutions (8-4) and (8-5), we transform the integral to

J

__ 1~2 ~z = j -i4z2 dz= j f (z)dz,

l + 3(z+z-') iz 3z4+10z +3

ct (o) 2 ct ( O) ct (O)

where f (z) = 3 ,a.; 1 ~;2+ 3. The singularities off are poles located at the points

where 3 (z^2 )

2

+ 10 (z^2 ) + 3 = O. Using the quadratic formula, we see that the