We
continue now looking at more examples of complex integration and application of
the residue theorem.
Example 1:
∫ -¥ ¥ (1 + x2 ) dx /
1 + x4
Find
the residues at exp(pi/4) and exp(3pi/4)
and hence integrate the preceding:
Solution:
f(z)
= 1 + x2 / 1 + x4
Now: We note that
1 + x4 = 0
when x = (-1) ¼
Or
the fourth root of (-1). The modulus of (-1) = 1 and the argument is p. Therefore,
by de Moivre’s theorem:
(-1 + 0i) ¼ = 4Ö(-1) [ cos (p + 2kp/ 4) + i sin (p + 2kp/ 4)]
k= 0, 1, 2, 3…….
At
k= 0: (-1) ¼ = cis (p/4)
= exp (pi/4)
At
k=1: (-1) ¼ = cis (3p/4)
= exp (3pi/4)
(We’re not concerned with k=2, 3……Why?)
We
therefore take the semi-circular contour as shown in the graphic. So that:
òCR f(z) dz + ∫ -R R f(x) dx
= 2 pi (sum
of residues inside C)
As R increases without bound we have:
∫ -¥ ¥ (1 + x2 ) dx /
1 + x4 =
2 pi [Res f(z) at exp (pi/4) and exp (3pi/4) ]
2 pi [Res f(z) at exp (pi/4) and exp (3pi/4) ]
For
the case f(z) = p(z)/ q(z) Then:
Res
f(z) z = a =
lim z ® a [p(z)/ q’(z)]
Then
for z = exp (pi/4):
Res
f(z) z = exp pi/4 =
lim z ® exp pi/4 1 + z2 /
4 z3 =
1
+ exp(pi/2) / 4 exp(3pi/4) = 1 + cis (p/2) / 4 cis (3p/4)
Res
f(z) z = exp pi/4 =
1 + i / 4(- Ö2/ 2 + iÖ2/ 2
)
= 1/ 2Ö2 (i+ 1/ i- 1)
Rationalize
the denominator to get: 2i/ -2 = - Ö2i/ 4
Then
for z = exp (3pi/4):
Res
f(z) z = exp 3pi/4 =
lim z ® exp 3pi/4 1 + z2 /
4 z3 =
1
+ exp(3pi/4) / 4 exp(9pi/4) =
¼
[exp(-3pi/4) +
exp(- 9pi/4)]
=
¼ [ cis (-3pi/4) + cis (- 9pi/4)]
¼ [ cis (-3pi/4) + cis (- 9pi/4)]
Therefore:
∫ -¥ ¥ (1 + x2 ) dx /
1 + x4 =
2 pi [Res f(z) at exp (pi/4) and exp (3pi/4) }
2 pi [Res f(z) at exp (pi/4) and exp (3pi/4) }
= 2 pi {- Ö2i/ 4 + ¼ [
cis (-3pi/4) + cis (- 9pi/4)]}
Problem
for Serious Math Mavens:
Obtain
the integral for:
∫ -¥ ¥ x dx /
(x2 - 2x +
2)
