Integrating a Complex Function: A Challenge

In summary, the author attempted to integrate sinc(x)^4 between negative infinity and infinity and found that the value of the residue in the lower contour is 2pi/3.
  • #1
Jukai
13
0

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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  • #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 definitely 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.
 

1. What is the definition of a complex function?

A complex function is a mathematical function that takes in a complex number as an input and outputs a complex number. It is typically written in the form f(z) = u(x,y) + iv(x,y), where z = x + iy, u and v are real-valued functions of x and y, and i is the imaginary unit equal to the square root of -1.

2. What is the difference between a real and a complex function?

The main difference between a real and a complex function is that a real function takes in a real number as an input and outputs a real number, while a complex function takes in a complex number as an input and outputs a complex number. Additionally, complex functions can exhibit more complex behavior than real functions due to the presence of an imaginary component.

3. Why is integrating a complex function considered a challenge?

Integrating a complex function is considered a challenge because it involves finding the antiderivative of a complex function. Unlike real-valued functions, complex functions have multiple ways of approaching integration, and the concept of a complex antiderivative is not as straightforward as in the real case. This can make integration of complex functions more difficult and require more advanced techniques.

4. What are some common techniques used to integrate complex functions?

Some common techniques used to integrate complex functions include contour integration, Cauchy's integral theorem, Cauchy's integral formula, and the residue theorem. These techniques involve manipulating and analyzing complex functions in order to find their antiderivatives.

5. How is integrating a complex function useful in real-world applications?

Integrating complex functions is useful in many real-world applications, especially in physics and engineering. For example, in electrical engineering, complex functions are used to model circuits and the integration of these functions can help determine important quantities such as voltage and current. In physics, complex functions are used to describe wave phenomena and integrating them can help calculate quantities such as energy and momentum. Additionally, complex functions and their integration are used in various areas of mathematics, such as complex analysis and differential equations.

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