Tricky calculus problem

[eskowitz2]

One of my coworkers asked me how to evaluate the integral of (sin(x)/x)^2 from -inf to inf and I am stumped. MATLAB says that the answer is pi, but I do not know how to solve it analytically.

Integration by parts does not work because I eventually get an integral with (cos(x))^2 in it and then I am essentially back where I started from. (And I can't take advantage of cos(x)^2 + sin(x)^2 = 1, because there is an integer multiple in front of cos(x)^2).

I tried using Euler's identity to break the sin(x) up into -j0.5e^(jx)+j0.5e^(-jx), but that lead me to an integral of sin(2x)/x, which I couldn't figure out either.

Apparently there is a class of integrals that I have totally forgotten how to solve. Any guidance would be appreciated.

Thanks,
-Mike

 

[Nathelion]

You couldn't just use an integral table? That's what I'd do... It's kind of "cheating" I suppose, but outside of very specific academic environments, people don't sit around doing calculus with pen and paper anymore...

 

[esun]

Not an easy thing to solve analytically (i.e. without a table).

Using the Integrator (http://integrals.wolfram.com/index.jsp) I get something in terms of Si (http://mathworld.wolfram.com/SineIntegral.html) which is not a closed form solution, so you likely cannot do this using any standard integral solving methods. Maybe a coordinate change would help.

 

[Biftheunderstudy]

I know I have done this at some point, but like you I have forgotten most of it. I would take a Laplace or Fourier transform though, since this is the approach used for sinx/x. So basically your looking for the inverse Laplace of 1/a(s^2+a^2).

 

[SocrPlyr]

The way to do it is definitely to take a Fourier Transform. The if you look at the definition of the transform if you plug in 0 for f, you get the integral you are looking for. So you can use the transform that is already computed for you in 203 in the table. Note with their definition of sinc, a=1/pi for your case. Thus the result is pi*triangle(0)=pi. If you actually need to prove all the steps in between, you are somewhat screwed b/c you have to prove the following:
1. The Fourier Transform is unique. (Hard but you can just say it is known for most proofs.)
2. The inverse transform of triangle is the sinc squared. (In conjunction with 1., this means that the forward transform is the triangle and you are done. I left out some constants, but you get the basic idea. This is not a very hard integral to compute, but it does require integration by parts and Euler's formula.)

Josh

 

[KillerCharlie]

This is screaming for the use of complex variables - actually, I believe that is what Fourier Transforms use.

 

[eskowitz2]

My coworker just sent me a solution to the problem. The solution involves using the trig identity (sin(x))^2 = (1-cos(2x))/2 and Euler's identity to rewrite the integral as

Re[ integral( 0.5 * (1-exp(2*j*x)) / x^2, x, -inf, inf ) ]

The rest of the solution involves performing a contour integral on a "half doughnut" in the complex plane. The half doughnut is basically a half circle with the (0,0) point notched out because the above equation has a first order pole at x=0.

I have never done contour integrals before so I will just quote the rest of the solution:



"Since the contour encloses no poles, this integral is zero. It has several pieces. As the large circle moves out to infinity, it contributes nothing to the integral because the 1/x^2 factor goes to zero rapidly enough. So the integral along the real axis (Iaxis) and the integral from going halfway around the pole (Ipole) combine to zero. So

Iaxis = -Ipole = pi*j cross residue at x = 0,

since going around the pole clockwise by an angle theta contributes theta cross residue, and here theta = pi.

The integrand's residue at x=0 is the coefficient of the 1/x term in infinite series expansion. Since 1 - exp(2*j*x) = -2*j*x + 2*x^2 + ..., the residue of (1-exp(2*j*x))/(2*x^2) at x = 0 is -j. So Iaxis is pi.

And Iaxis is the original integral so integral( (sin(x)/x)^2, x, -inf, inf ) = pi."



This doesn't make much sense to me, but maybe it does to you...

-Mike

 

[Biftheunderstudy]

Unfortunately, yes it does make sense. Unfortunately because I tried ever so hard to forget residue calculus and contour integrals in complex space....

 

[KillerCharlie]

Hah, I knew it...

...but that's because I'm learning it right now in class...

 

[eskowitz2]

What course would these topics be covered in? I've had a lot of math, but I've never worked with contour integrals before.

 

[SocrPlyr]

I'm still unsure why you would go through all of that instead of just using the FT and a lookup table which is readily available.

Josh

 

[Born2bwire]

Originally posted by: eskowitz2
My coworker just sent me a solution to the problem. The solution involves using the trig identity (sin(x))^2 = (1-cos(2x))/2 and Euler's identity to rewrite the integral as

Re[ integral( 0.5 * (1-exp(2*j*x)) / x^2, x, -inf, inf ) ]

The rest of the solution involves performing a contour integral on a "half doughnut" in the complex plane. The half doughnut is basically a half circle with the (0,0) point notched out because the above equation has a first order pole at x=0.

I have never done contour integrals before so I will just quote the rest of the solution:



"Since the contour encloses no poles, this integral is zero. It has several pieces. As the large circle moves out to infinity, it contributes nothing to the integral because the 1/x^2 factor goes to zero rapidly enough. So the integral along the real axis (Iaxis) and the integral from going halfway around the pole (Ipole) combine to zero. So

Iaxis = -Ipole = pi*j cross residue at x = 0,

since going around the pole clockwise by an angle theta contributes theta cross residue, and here theta = pi.

The integrand's residue at x=0 is the coefficient of the 1/x term in infinite series expansion. Since 1 - exp(2*j*x) = -2*j*x + 2*x^2 + ..., the residue of (1-exp(2*j*x))/(2*x^2) at x = 0 is -j. So Iaxis is pi.

And Iaxis is the original integral so integral( (sin(x)/x)^2, x, -inf, inf ) = pi."



This doesn't make much sense to me, but maybe it does to you...

-Mike

This would be something you would cover in a complex variables class.

 

[Biftheunderstudy]

For me it was in a 3rd year course called mathematical physics. It dealt with complex variable calculus, power series solutions to differential equations, bessel and similar functions, and a few other nasty surprises. Basically its the calculus you need to do the more advanced E&M (retarded solutions, greene's functions etc.), Quantum Mechanics etc. E&M actually covers a few bases as well since the math is very similar to Relativity and fluid dynamics.

 

[Born2bwire]

Originally posted by: Biftheunderstudy
For me it was in a 3rd year course called mathematical physics. It dealt with complex variable calculus, power series solutions to differential equations, bessel and similar functions, and a few other nasty surprises. Basically its the calculus you need to do the more advanced E&M (retarded solutions, greene's functions etc.), Quantum Mechanics etc. E&M actually covers a few bases as well since the math is very similar to Relativity and fluid dynamics.

Yeah, I use it a lot in E&M with Sommerfeld integrals and such.

 

[Special K]

Originally posted by: Biftheunderstudy
I know I have done this at some point, but like you I have forgotten most of it. I would take a Laplace or Fourier transform though, since this is the approach used for sinx/x. So basically your looking for the inverse Laplace of 1/a(s^2+a^2).

Wait, are you saying that sin(x)/x has an indefinite integral if you use Laplace transforms? I thought sin(x)/x belonged to a class of functions that did not have indefinite integrals? (what was the name of those functions? I can't seem to remember).

 

[soccerballtux]

Originally posted by: Special K

Originally posted by: Biftheunderstudy
I know I have done this at some point, but like you I have forgotten most of it. I would take a Laplace or Fourier transform though, since this is the approach used for sinx/x. So basically your looking for the inverse Laplace of 1/a(s^2+a^2).

Wait, are you saying that sin(x)/x has an indefinite integral if you use Laplace transforms? I thought sin(x)/x belonged to a class of functions that did not have indefinite integrals? (what was the name of those functions? I can't seem to remember).

My DSP class was a year ago but it definitely has something to do with the sync function, and Fourier Transforms.

Can't remember either heh.

edit: actually, I'm going to take this back; as after a short wiki I am clearly talking out of my butt. Sorry ^_^