Contour integration on sin^2x/x

[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.)

Homework Statement

Prove
$$I = \int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2} \,dx=\pi.$$

Homework Equations

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.

 

[Nino MMP]

The integral is not zero, since the function $\frac{\sin^2(z)}{z^2}$ is not holomorphic. We can evaluate the integral using contour integration by integrating along a rectangular contour with vertices at $-R, R, R+i\pi, -R+i\pi$ in the upper half plane. Then the integral around the boundary of this contour is\begin{align*}\int_{-R}^R \frac{\sin^2 x}{x^2}\, d x + \int_R^{R+i\pi} \frac{\sin^2 z}{z^2} \,dz + \int_{R+i\pi}^{-R+i\pi} \frac{\sin^2 z}{z^2} \,dz + \int_{-R+i\pi}^{-R} \frac{\sin^2 x}{x^2}\, d x \end{align*}The first and fourth integrals are equal, since they are just the integral from $-R$ to $R$ along the real axis. The second and third integrals are also equal, since the integrands are even functions in the complex plane and we are integrating over the same length. We can therefore simplify the integral around the boundary to \begin{align*}2 \int_{-R}^R \frac{\sin^2 x}{x^2}\, d x + 2 \int_R^{R+i\pi} \frac{\sin^2 z}{z^2} \,dz\end{align*}Using the fact that $\sin^2 (z) = \frac{1}{2} (e^{2iz} + e^{-2iz})$, we can rewrite the second integral as\begin{align*}\frac{1}{2} \int_R^{R+i\pi} \frac{e^{2iz} + e^{-2iz}}{z^2} \,dz\end{align*}We can now use Cauchy's theorem to evaluate this integral, since the integrand is holomorphic on the interior of the contour. Thus, the integral is zero. Therefore the integral around the