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 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:

ò_CR   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!)