## Monday, January 27, 2014

### More Complex Integrals

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:
Begin by identifying the singular points where:
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 pTherefore, 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) ]
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)]

Therefore:
-¥  ¥  (1 + x2 )  dx / 1 + x4     =

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)
(Hint: Remember  the residue formula (i.e. for a given pole of order m) and your complex numbers!)