# Solving int (sin(x)/x)^2 dx

1. Sep 20, 2009

### Susanne217

1. The problem statement, all variables and given/known data

Hi I am working on a solution for the integral the integral

$$\int_{-\infty}^{\infty} (\frac{sin(x)}{x})^2 dx$$

2. Relevant equations

3. The attempt at a solution

I know from theory that

$$\int_{-\infty}^{\infty} f(x) dx = \int_{-\infty}^{a} f(x) dx + \int_{-\infty}^{a} f(x) dx$$

So that $$\int_{-\infty}^{1} (\frac{sin(x)}{x})^2 dx + \int_{1}^{\infty} (\frac{sin(x)}{x})^2 dx$$

since $$(\frac{sin(x)}{x})^2 = \frac{sin^{2} x}{x^2}$$

and by trig identity

$$\frac{sin^{2} x}{x^2} = \frac{1-cos(2x)}{2} \cdot x^{-2}$$

for clarity that gives me an integral that I need to solve where

$$\int_{-t}^{t} \frac{1-cos(2x)}{2x^2} dx = \int^{1}_{-t}\frac{1-cos(2x)}{2x^2} dx + \int^{t}_{1}\frac{1-cos(2x)}{2x^2} dx$$

What I am simply am burning to know is that the correct approach?

I have read in another thread in this forum that the purpose of the task is to first find a corresponding taylor series expansion

so anyway since $$\frac{sin^2 x}{x^2} = \frac{1-cos(2x)}{2x^2} = \frac{-(cos(2))-1}{2}\cdot (x-1) - \frac{cos(2) - 4 \cdot sin(2) + 3}{2} \cdot (x-1)^2 + 2 \cdot (7 sin(2) -6) \cdot (x-1)^3 + \cdot + ???$$

I used $$x_0 = 1$$ since there is hole x = 0

Last edited: Sep 20, 2009
2. Sep 20, 2009

### snipez90

I think you have some good ideas, and some that might not work. For instance, I don't know which thread you're referring to about the taylor series expansion, but it sounds like a messy approach.

Now if you already know the value for $\int_{0}^{\infty}\frac{\sin x}{x}\,dx,$ then this is even easier to evaluate. But that approach is probably not what you want, but we can do something else.

First of all, the integrand is even, so by parity symmetry,

$$\int_{-\infty}^{\infty}\frac{\sin ^2 x}{x^2}\,dx = 2\int_{0}^{\infty}\frac{\sin ^2 x}{x^2}\,dx.$$

Convergence is not really an issue, since we have the x^2 in the denominator. By the trig identity you wrote down, the integral becomes

$$\int_{0}^{\infty}\frac{1-cos(2x)}{x^2} \,dx.$$

Now there is a standard trick for writing the term 1/x^2 as the integral of an exponential as follows:

$$\frac{1}{x^{2}}=\int_{0}^{\infty}te^{-tx}\,dt.$$

This is easy to verify by basic calculus. We can then work with a double integral by writing

$$\int_{0}^{\infty}\frac{1-cos(2x)}{x^2} \,dx = \int_{0}^{\infty}\int_{0}^{\infty}te^{-tx}(1-\cos 2x)\,dt\,dx .$$

What we want is to switch the order of integration (i.e. switching the places of dt and dx and integrate with respect to x). This is justified by Fubini's theorem and Tonelli's theorem. The rest is more antiderivative computation from basic calculus.

Note that the same approach can be used to determine the value of $\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx,$ but showing convergence is trickier. There is probably a way to evaluate your integral via fourier transforms and working with distributions, but the approach I outlined is completely elementary.

3. Sep 20, 2009

### Susanne217

So the trick is to work out that integral?

But if I try to work out the that integral I get undefined on my calculator. So what do I need to do here?

I can see that I can return it to the original integral. Is that what I need to do ?

Sincerely

Susanne

P.s. I know that according to parsevals identity that original integral $$\int_{0}^{\infty}\frac{sin^2 x}{x^2} = \frac{\pi}{2}$$ then our integral must be $$\frac{\pi}$$?

Last edited: Sep 20, 2009
4. Sep 20, 2009

### snipez90

Don't use a calculator for actual math problems.

After switching the order of integration, we have to evaluate

$$\int_{0}^{\infty}\frac{1-cos(2x)}{x^2} \,dx = \int_{0}^{\infty}\int_{0}^{\infty}te^{-tx} (1-\cos 2x)\,dx\,dt.$$

In other words, we treat t as a constant and integrate with respect to x (inner integral first of course). If you're having trouble with integrating

$$te^{-tx} (1-\cos 2x)$$

with respect to x (t constant), you could always ask for help or use wolfram's integrator as a last resort.

5. Sep 20, 2009

### Susanne217

sorry

then I get $$\lim_{t \to \infty } \int_{0}^{t} \frac{e^{-xt}((cos(2x)\cdot t^2 -t^2 - sin(2x)t-4)}{t^2+4} dt = \pi$$ how is that?

Last edited: Sep 20, 2009
6. Sep 20, 2009

### snipez90

No, sorry it was a latex error, the (1-cos(2x)) is not part of the exponential. Incidentally, the value of the integral you posed is $\pi,$ which is also what you get for $\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx.$ However, you do have to demonstrate the relationship between the two, which can be done via differentiation under the integral sign or I think even a simple integration by parts.

7. Sep 20, 2009

### Susanne217

I have done it hopefully correctly in my previous post. Hopefully it was what you were thinking of :)

8. Sep 20, 2009

### snipez90

Um, it looks like you got the inner integral for the most part, except you are missing a factor of 2 on the sin(2x)t term. Then you should have

$$\int_{0}^{\infty}\int_{0}^{\infty}te^{-tx} (1-\cos 2x)\,dx\,dt = \int_{0}^{\infty}\left[ \frac{e^{-xt}((cos(2x)\cdot t^2 -t^2 - 2sin(2x)t-4)}{t^2+4} \,dt \right]_{x = 0}^{x = \infty}\,dt.$$

As x goes off to infinity, that big expression vanishes. Evaluate the big expression at x = 0 (don't forget the negative sign), then manipulate the result to make use of $\int \frac{1}{1+t^2}\,dt = \arctan t + C$ and you should be done.

Last edited: Sep 20, 2009