Integral using Residue Theorem

In summary, the integral of 1/(1+x^4) from -infinity to +infinity is 2pi*i*sum of residues at z=-sqrt(i). The Attempt at a Solution states that z^4=-1 and can be found by first letting z=r e^{j\theta} and solving for z^4. The residue theorem states that residue at z=-sqrt(i) is 1/(4*i*sqrt(i)) and at z=i*sqrt(i) is 1/(4*sqrt(i)). When z^4+1=0, there are 4 unique singularities for n=0, 1, 2
  • #1
yeahhyeahyeah
30
0

Homework Statement



the integral of 1/(1+x^4) from -infinity to +infinity


Homework Equations




Residue theorem.

The Attempt at a Solution



1/(1+z^4) so z^4 = -1

I know I should be using the residues at z = -sqrt(i) and z= i*sqrt(i)

I am getting a complex number as an answer which makes no sense

residue at z = -sqrt(i) = 1/(4*i*sqrt(i))
and at z = i*sqrt(i) = 1/(4*sqrt(i))

and therefore integral of (1/1+z^4) = 2pi*i* sum of those residues


Am I on the right track?
 
Physics news on Phys.org
  • #2
This is a tricky. When else does [tex] z^4+1 =0 [/tex]? Maybe at [tex] z=e^{i\frac{\pi}{4}} [/tex].
Can you get it from here?
 
Last edited:
  • #3
been a while since I've done these, but as a start, rather than working with sqrt i, which i don't think is unique, I would find the singularities by first letting
[tex] z = r e^{j \theta} [/tex]

whilst for some integer n
[tex] -1 = e^{j\pi(2n+1)} [/tex]

then the singularities can be found by
[tex] z^4 = r^4 e^{j 4 \theta} = e^{j\pi(2n+1)}[/tex]

giving
[tex] \theta = \frac{\pi(2n+1)}{4} = \pi(n/2+1/4)[/tex]

which I think will give 4 unique singularities for n = 0, 1, 2, 3
 
  • #4
OOOh, I see, I used the singularities and I get pi/sqrt2 which I think is right. Thank you guys so much. I actually have another related question now though. First of all, why does sqrt(i) not work when I use it as z?

Also,

I'm now trying to do:

integral (0 to infinity) of sin(x^2) using residue theorem as well. I can't figure out how to make the substitution.


I don't know when/if you guys will answer that question but thanks so much for your help just now!
 
  • #5
well first you didn't find all the singularities (2 out of 4) & sqrt(i) is not unique. There are 2 unique answers, note we try and solve
[tex] z^2 = i [/tex]

so as before
[tex] (e^{i \theta})^2 = e^{i 2 \theta} = e^{(2n+1)\pi} [/tex]

so the 2 solutions in the \theta range [0, 2pi) are
[tex]\theta = (n+1/2)\pi} = \pi/2, 3\pi/2 [/tex]

compare it with solving
[tex] z^2 = 1 [/tex]
there is a plus & minus solution

that said if i remember residues correctly, you probably only need 2 of them anyway

not too sure about your sin question, but it may be worth trying writing the sin in terms of the sum/differnce of 2 complex exponentials
 
  • #6
lanedance said:
been a while since I've done these, but as a start, rather than working with sqrt i, which i don't think is unique, I would find the singularities by first letting
[tex] z = r e^{j \theta} [/tex]

whilst for some integer n
[tex] -1 = e^{j\pi(2n+1)} [/tex]

then the singularities can be found by
[tex] z^4 = r^4 e^{j 4 \theta} = e^{j\pi(2n+1)}[/tex]

giving
[tex] \theta = \frac{\pi(2n+1)}{4} = \pi(n/2+1/4)[/tex]

which I think will give 4 unique singularities for n = 0, 1, 2, 3
I agree. Except when you make the semicircle, the poles n=0, 1 are the only ones on the upper half plane.
 

What is the Residue Theorem?

The Residue Theorem is a powerful tool in complex analysis that allows for the evaluation of certain integrals using the residues of a function. It states that the integral of a function over a closed contour can be expressed as the sum of the residues of the function at its singularities within the contour.

How is the Residue Theorem used in evaluating integrals?

The Residue Theorem is used to evaluate integrals by first identifying the singularities of the function within the closed contour. Then, the residues at these singularities are calculated and added together to obtain the value of the integral. This method is particularly useful for integrals that cannot be evaluated using traditional techniques.

What are the benefits of using the Residue Theorem?

The Residue Theorem allows for the evaluation of complex integrals that would otherwise be difficult or impossible to solve. It also provides a systematic and efficient approach to evaluating these integrals, making it a valuable tool for scientists and mathematicians.

Are there any limitations to using the Residue Theorem?

While the Residue Theorem is a powerful tool in complex analysis, it does have some limitations. It can only be used for integrals over closed contours, and the function being integrated must have a finite number of singularities within the contour.

How is the Residue Theorem related to complex functions and analyticity?

The Residue Theorem is closely related to the concept of analyticity in complex functions. A function is considered analytic if it is differentiable at every point in its domain. The Residue Theorem relies on the fact that analytic functions can be written as a sum of their singularities, which allows for the use of residues in evaluating integrals.

Similar threads

  • Advanced Physics Homework Help
Replies
19
Views
700
  • Calculus and Beyond Homework Help
Replies
23
Views
1K
  • Advanced Physics Homework Help
Replies
15
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
810
  • Advanced Physics Homework Help
Replies
2
Views
3K
Replies
7
Views
1K
Replies
4
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
765
Back
Top