Integrating sinc(x)^4 between negative infinity to infinity using complex analysis

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  • #1

Homework Statement

Integrate the following:

(sin(x)/x)^4 between negative infinity and infinity.

Homework Equations

The residue theorem, contour integral techniques.
The answer should be 2pi/3

The Attempt at a Solution

I'm not even sure where to start honestly. I define a function f(z)=(sin(z)/z)^4. I'm not quite sure what to make of the point z=0, but I make a contour integral the shape of half a donut in the upper half plane with a little half-circle above z=0. So, I have 3 integrals to consider, the principal value integral on the x-axis, the one on the little half-circle and the big half-circle.

According to the residue theorem, the sum of these integrals should give me 0. I'm pretty sure that using Jordan's lemma, we can prove that the integral on the big half-circle is 0. Also, the principal value integral on the x-axis is the original function. What do I do with the last part now?

I define z=εe^(iθ) there and insert in my function. The integral is between pi and 0, and I need to take the limit of ε as it goes to 0.

I'm honestly lost, is there any chance someone could help me at least start this problem? I don't know if what I've written above is correct or not. Just a little help please =(, this problem has been a great a source of stress for me recently.
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Answers and Replies

  • #2

You should use the relation
[tex] \sin^4\theta = \frac{3 - 4 \cos 2\theta + \cos 4\theta}{8}[/tex]
It will make your life much easier ^^
  • #3

What you have written above is absolutely correct. You have 4 paths, let's call them P1,P2,P3,P4.
P1 goes from -infinity to -epsilon
P2 is a small semicircle from -epsilon to epsilon
P3 goes from epsilon to infinity
P4 is the large semicirle
We already know that the integral over all paths combined is 0.
Also integral over P4 is 0.
This tells us that
[tex] \int_{P1, P3} f(z) dz = - \int_{P2} f(z) dz [/tex]
Obviously, the combined path P1, P3 is "almost" what we are looking for.
Now chose
[tex] f(z) = \frac{1}{8} \frac{3-4e^{2iz}+e^{4iz}}{z^4} [/tex]
The REAL part of [itex]f(z)[/itex] is the function that we want to integrate.
Now you need to parametrize P2. choose
[tex] z = \epsilon e^{i \theta} [/tex]
and integrate over [itex]\theta\in[0,\pi][/itex] while [itex]\epsilon \rightarrow 0 [/itex]
L'Hôpital might come in handy in your calculations.
You might get a complex number as a result. Its real part will be the result you are looking for.
Let me know if anything remains unclear.
  • #4

Thank you for the reply.

I will try integrating with the relation you've proposed.

As for the other method, the one I started with, the limit tends to infinity in my calculations.

If z=εe^(iθ), then dz=iεe^(iθ) and the equation you've written becomes

[tex] f(z) = \frac{1}{8} \frac{(3-4e^{2iεe^{iθ}}+e^{4iεe^{iθ}})iεe^{iθ}}{(εe^{iθ})^4} [/tex]

You can factorize your epsilon above to have ε^3 in the bottom. Now you have a form of 0/0, which is alright because you can use L'Hôpital's. After that however, we have something that tends to infinity as ε-->0.

[tex] f(z) = \frac{1}{8} \frac{(-8ie^{iθ}e^{2iεe^{iθ}}+4ie^{iθ}e^{4iεe^{iθ}})ie^{iθ}}{4ε^{3}(e^{iθ})^4} [/tex]
  • #5

Try l'Hopital with respect to z, because we're in the region close to 0
  • #6

Hello, thanks for the help a few days ago, I found a way to solve this problem and I came back to give some closure. If we use sin(x)=(e^ix-e^-ix)/2i, and multiply that by itself four times (to the power 4), we will get some positive and some negative exponentials. There is a pole at z=0. We need to consider 2 contours, one for the positive exponentials and one for the negative exponentials.

For the positive exponentials, the contour is above the real axis and goes around the pole at z=0. Since there is no residue in here, the integral is 0.

For the negative exponentials, the contour is below the real axis and has the a residue inside. Therefore, to compute the integral of sinc(x)^4 between negative infinity and infinity, we just need to find the value of the residue in the lower contour. Using the formula for a residue of a pole of order 4:

Residue=-2ipi * lim(z->0) of (1/((4-1)!))*(d³/dz³)(((z-0)^4)*(negative exponentials))

We find that the integral is equal to 2pi/3.
  • #7

Alright, yeah, sounds reasonable and definately easier.
The thing is when you look up how to integrate sinc(x), you usually stumble across the contour integration method, because it's a good example for how to use different paths to obtain the value in an indirect way. However, there is usually always a small note at the bottom saying that you could just as well compute the residue at z=0 directly lol

Anyhow, sorry I couldn't be more helpful on that one. Glad you figured it out.

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