# Contour integration on sin^2x/x

1. Feb 24, 2010

### fantispug

SOLVED. (Sorry I worked out the solution when I was typing it and I can't work out how to remove the post; for some reason I assumed sin(z)<1 for complex z which is very false.)

1. The problem statement, all variables and given/known data
Prove
$$I = \int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2} \,dx=\pi.$$

2. Relevant equations

3. The attempt at a solution
For a first attempt I tried to integrate
$$\int_{0}^{\infty} \frac{\sin^2(\beta x)}{x^2} e^{-\alpha} \,dx$$ by differentiating twice under the integral, then integrating twice with respect to $$\alpha$$. I wasn't very careful with the branches of my logarithms and got an answer for the integral $$I = i \ln(-1)$$; so if I was more careful I think I could prove it that way.

However I though this would be a good integral to evaluate by contour integration; integrating along the semi-circle lying on the upper half plane, centre the origin, radius R anti-clockwise (then take the limit as R goes to infinity).

Now $$\frac{\sin^2(z)}{z^2}$$ is holomorphic, so the integral around this contour must be zero. Thus we get
$$\int_{-R}^{R} \frac{\sin^2(x)}{x^2} \dx=- \int_{0}^{\pi} \frac{sin^2(R e^{i\theta})}{(Re^{i\theta})^2} (iRe^{i\theta} \,d \theta)$$
So it follows
$$\left| \int_{-R}^{R} \frac{\sin^2(x)}{x^2} \dx \right | <= \frac{\pi}{R}$$.
Thus the integral I is zero.
What have I done wrong here? I am certain the function is holomorphic, and it makes sense the integral along the boundary should decay at infinity - why am I getting the wrong value for the integral?

Thanks.

Last edited: Feb 24, 2010