Integration by Residue Calculus

In summary, the conversation discussed evaluating the integral I(a)=∫cos(ax)/(x^4+1) from 0 to ∞ by considering the complex integral J=∮e^(iaz)/(z^4+1) over a semi-circle in the upper half plane and then taking the real part. The residue calculus method was used by evaluating the residues at the simple poles inside the semi-circle, but there was difficulty in computing the residues. The final answer, J=π√2/2e^(-a√2/2)(cos(a√2/2)+sin(a√2/2)), was obtained for the case of a>0.
  • #1
stephenkeiths
54
0

Homework Statement


I'm trying to evaluate the integral

[itex]I(a)=\int\frac{cos(ax)}{x^{4}+1}[/itex]
from 0 to ∞

Homework Equations



To do this I'm going to consider the complex integral:

[itex]J=\oint\frac{e^{iaz}}{z^{4}+1}[/itex]

Over a semi-circle of radius R in the upper half plane, then let R-->∞

There are 2 curves, the straight curve along the real axis, and the semi-circular arc.

The Attempt at a Solution


Along the first curve (along the real axis) the line integral is just

[itex]I_{1}=\int\frac{e^{iax}}{x^{4}+1}[/itex]

from 0 to infinity

Now the real part of this is 2*I(a)

The second line integral goes to zero. So 2*I(a)=Re(J).

So I need to evaluate J by residue calculus and take the real part to get I(a).

So there are 2 simple poles inside of the semi-circle, namely
[itex]z=e^{i\frac{\pi}{4}}[/itex] and [itex]z=e^{i\frac{3\pi}{4}}[/itex]

So [itex]J=2i\pi(Res(f,e^{i\frac{\pi}{4}})+Res(f,e^{i\frac{3\pi}{4}}))[/itex]

But I'm having trouble computing the Residues.

I know that [itex]Res(f,e^{i\frac{\pi}{4}})=\frac{e^{iaz}}{4z^{3}} z=e^{i\frac{\pi}{4}}[/itex]

But I'm struggling with the exponential to an exponential, I can't get a "clean" answer. I'm not sure if there's a better way I should go about this. Or if there's something I'm doing wrong.

Please Help!
 
Physics news on Phys.org
  • #2
stephenkeiths said:
But I'm having trouble computing the Residues.

I know that [itex]Res(f,e^{i\frac{\pi}{4}})=\frac{e^{iaz}}{4z^{3}} z=e^{i\frac{\pi}{4}}[/itex]

Are you sure about that result? I don't think there should be a z in the numerator. It looks like you calculated the residue using L'Hopital's rule, is that right? i.e.,

$$\lim_{z\rightarrow z_1} (z-z_1)\frac{e^{i\alpha z}}{z^4+1} = \lim_{z\rightarrow z_1} \frac{(z-z_1)'}{(z^4+1)'}e^{i\alpha z}.$$
When you apply the rule, you will get ##4z_1^3## in the denominator, but your numerator should just be a constant.

But I'm struggling with the exponential to an exponential, I can't get a "clean" answer. I'm not sure if there's a better way I should go about this. Or if there's something I'm doing wrong.

Please Help!

You can always write the roots in cartesian form, ##z = a + ib##.
 
  • #3
Thanks for replying

Thats what I got for my residue, Sorry that z is just saying plug in e^i...

I try writing it in that form (a+bi), but its so messy!

What I get is

[itex]J=\frac{\pi\sqrt{2}}{2}e^{-a\frac{\sqrt{2}}{2}}(cos(a\frac{\sqrt{2}}{2})+sin(a\frac{\sqrt{2}}{2}))[/itex]

Does anyone else think I'm crazy? I really feel like somethings wrong here!
 
  • #4
stephenkeiths said:
Thanks for replying

Thats what I got for my residue, Sorry that z is just saying plug in e^i...

I try writing it in that form (a+bi), but its so messy!

What I get is

[itex]J=\frac{\pi\sqrt{2}}{2}e^{-a\frac{\sqrt{2}}{2}}(cos(a\frac{\sqrt{2}}{2})+sin(a\frac{\sqrt{2}}{2}))[/itex]

Does anyone else think I'm crazy? I really feel like somethings wrong here!

If that's your answer for the whole integral, I'll agree with it, at least for the case of a>=0. If you aren't give that, I'd write |a| instead of a. I'll agree it's not a "really clean" answer. But it's about as clean as you are going to get.
 
  • #5
Ya I only need to solve for a>0

Thanks! It gives me some confidence!
 

1. What is Integration by Residue Calculus?

Integration by Residue Calculus is a method used to evaluate complex integrals that cannot be solved using traditional techniques. It involves using the concept of residues, which are complex numbers associated with singularities of a function, to find the value of the integral.

2. When is Integration by Residue Calculus used?

Integration by Residue Calculus is typically used when the integrand (the function being integrated) has singularities, such as poles or branch points, that make it difficult to integrate using other methods.

3. How does Integration by Residue Calculus work?

The method involves finding the residues of the function at its singularities, then using the Residue Theorem to evaluate the integral. This theorem states that the integral of a function around a closed contour is equal to the sum of the residues of the function inside the contour.

4. What are the benefits of using Integration by Residue Calculus?

Integration by Residue Calculus allows for the evaluation of complex integrals that would otherwise be impossible to solve. It also provides a faster and more efficient method for solving certain types of integrals, compared to traditional techniques.

5. Are there any limitations of Integration by Residue Calculus?

One limitation of this method is that it can only be used for functions with singularities. It also requires a good understanding of complex analysis and the Residue Theorem, which can be challenging for some individuals.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
356
  • Calculus and Beyond Homework Help
Replies
2
Views
812
  • Calculus and Beyond Homework Help
Replies
1
Views
492
  • Calculus and Beyond Homework Help
Replies
16
Views
475
  • Calculus and Beyond Homework Help
Replies
3
Views
812
  • Calculus and Beyond Homework Help
Replies
2
Views
904
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
269
Back
Top