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!)

More Laurent Series Examples

No, we’re not yet finished with these exotic mathematical denizens! We want to give yet more examples, and especially showing different ways to obtain the Laurent equivalent of different functions.

 

Ex. (1):   exp(2z)/ (z - 1)³  about  z ₀  =   1

 

Let u = (z – 1)  so that z = (u -1)

Then:

 exp ( 2z) / (z - 1)³     =  exp(2 + 2u) /  u ³  =   [( e) ²  /  u ³   ] /  e ^2u 

 

=   e ²  /  u ³      [ 1  + 2u   + (2u) ² / 2!   +   (2u) ³ / 3!   + 

 (2u) ⁴ / 4! +     …….   

=  e ²  /  (z - 1)³     +   2 e ²  /  (z - 1)²      +   2 e ²  /  (z – 1)  +  

4 e ²  / 3  +  2 e ²  / 3 (z – 1)  +   …….

 

Note that z is a pole of order 3 or a “triple pole” (Why?)    Note also the series converges for all values of z  ¹  1

Ex. (2):   f(z) =  z – sin z /  z ³

We’ve already seen the series for sin z:

 

sin(z) = z -  z³ / 3!  +  z⁵ / 5!  …..  -  (-1) ^{n- 1} z ^{2n -1}/  (2n -1)!  +

 

Then:

z – sin z /  z ³     =  1/  z ³    [z -  (z -  z³ / 3!  +  z⁵ / 5!  …..  ) ] 

 

 =  1/  z ³    [z³ / 3!   -  z⁵ / 5!    +   z ⁷ / 7!  -    ……….

 

 

=   1/3!  -  z ² / 5!     +   z ⁴ / 7!  -    z ⁶ / 9!   + ……..

 

Ex. (3)

 Find a Laurent series for f(z) =  sin (4z) / z ⁴

 

Using the sine series we can write:

 

sin (4z) / z ⁴     =  1/   z ⁴   [4z -    4z³ / 3!  +  4 z⁵ / 5!  +  ……..]

 

=     4 /   z ³    -    4³ / 3!  z  +   4⁵  z  / 5!       +   …..

 

= 4 z  -³    -  4³  z -¹ /  3!   +   4⁵  z  / 5!        +   …..

 

 

 

Which on inspection is found to have the form:

 

å^¥ _{n = 0}    (-1) ⁿ   4 ^{2n+ 1}  z ^{2n -3}  /   (2n +  1)!

Check term number 3 (n = 2):

 

4 ^{2n+ 1}     = 4 ^{2(2)+ 1}      =      4⁵

 

z ^{2n -3}  =  z ^{2(2) - 3}      = z ¹=  z

 

(2n +  1)!   =  (2(2) +  1)!  = 5!

 

Then:  we get  4⁵  z  / 5!        

 

A removable singularity occurs at z = 0 so we expect the region of convergence to be:

 

÷ z  ÷      >   0   

 

Problems for Math Mavens:

 

Let f(z) = exp (-1/z ²) /  z⁵

 

a) Show the Laurent series can be written: 

 

å^¥ _{n = 0}    (-1) ⁿ   /   n !  z ^{2n + 5}   

b)Specify where the singularity would occur and the precise region of convergence

 

 

Challenge Problem: For über Math Mavens only –

 

Let f(z) = 7 z ²  + 9 z  - 18 /  z ³ -  9z

 

Find Laurent series for the convergence regions:

 

a)  0  < ÷ z ÷   <    3  and

b)    ÷ z  ÷      >   3  

 

(Hint: Approach the function f(z) by resort to partial fraction decomposition)

 

 

More on Laurent Series

Note to readers: I reiterate once more that this set of blog posts isn’t intended to be comprehensive nor should they be seen as replacing the hard work of diligently working problems on your own using available text books. They are merely intended to spur on curiosity and perhaps offer a few different perspectives from what one might have seen already.

Included here are the coefficients in typical Laurent series which are generally obtained by other means (i.e. than appealing directly to integral representations). Some examples I have included below might suffice to expose this.

 

Ex. (1):  You have the Maclaurin series:

 

exp (z) = å^¥ _{n = 0}   (z) ⁿ  / n!  = 1 + z + z² / 2!  + z³ / 3!  +…..

 

 

Now, replace z by 1/z in the expansion:

 

 

exp( 1/z) = å^¥ _{n = 0}   1  / n!   z ⁿ  = 1 + 1/ 1! z   + 1/ 2!  z² 

 +  1/ 3! z³   +…..

 

for: 0   <   ÷ z ÷    <     ¥ 

 

Which then becomes a Laurent series expansion.

 

To establish this on inspection note that no positive powers of z appear, only negative, i.e. then 1/z is 

z ^{- 1}   

 So, in effect we can say that the coefficients of the positive powers are zero.  It’s also important to note here that the coefficient of : 

 

1/ 1! z     = 1/ z

 

is unity, so according to Laurent’s theorem we designate that the coefficient:

 

c _n  =  1/2 pi  ò_C  exp( 1/z)  dz

 

where C is any positively oriented simple closed contour around the origin.  Now, since from our prior  Dec. 12, Introducing Laurent Series) examination c _n  =  1 then:
 

1  =1/2 pi  ò_C  exp( 1/z)  dz   and:  ò_C  exp( 1/z)  dz    =  2 pi 

Ex. (2):  Consider the function:

f(z) = 1/ (z – i) ²

 

This is already in the form of a Laurent series , where z ₀  =   i.  That is,

 

å^¥ _{n = -¥}   c _n (z – i) ⁿ     (For:  0   <   ÷ z - i ÷    <     ¥  )

Where in this case,  c _{- 2}  =  1  and all other coefficients are zero.  Then from the previous theorems, relations we’ve explored (blog post of December 12, ‘Introduction to Laurent Series’, sub-header: ‘More intricacies and singularities’):

 

c _n  =  1/2 pi  ò_C   dz/ ( z – i) ^{n +3}      n = 0, +1, +2, +3 …..

 

 

where C denotes any positively oriented circle ÷ z - i ÷   = R  about the point: z ₀  =   i

 

 

Then it follows from this:

 

 

a)    ò_C   dz/ ( z – i) ^{n +3}      =  0   (when n ¹   2)

 

b) ò_C   dz/ ( z – i) ^{n +3}      =  1    (when n=    2)

 

Problems for the Math Maven:

 

1)     The function: f(z) =  -1/ (z – 1) (z – 2)
 

a) Rewrite it using the partial fraction form

 

b) Identify the two singular points.

 

 

c) If  the function is analytic in the domains:

 

÷ z ÷    <   1;   1     < ÷ z ÷     <      2   and:   2    <  ÷ z ÷    <     ¥,  

 

 

Draw a sketch showing the different domains.

 

 

2) Find the Laurent series that represents the function:

 

 

f(z) =  z²  sin (1 / z² )

 

In the domain:  : 0   <  ÷ z ÷   <     ¥ 

 

 

Hint:  Recall the series for sin(z) = z -  z³ / 3!  +  z⁵ / 5!  …..-  

   (-1) ^{n- 1} z ^{2n -1}/  (2n -1)!  +   ……   (For ÷ z ÷   <    ¥  )