Residue theorem applied to $\int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2}$

Question

I am trying to use residue theorem applied to $\int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2}$ over an indented contour. I am also told to consider integrating $\frac{e^{\pm2iz}-1}{z^2}$. I chose a bridge shape contour where we have $\Gamma_1:R+iy;0\leq y; \Gamma_2:x+iR.\Gamma_3=-\Gamma_1. \Gamma_{\epsilon}:\epsilon e^{2i\theta}$. I was able to show that $\Gamma_1,\Gamma_2,\Gamma_3$ go to $0$ as $R \to \infty$. For the $\Gamma_{\epsilon}$ I have tried this so far: $$\int_{\Gamma_{\epsilon}} \frac{e^{2iz}}{z^2}dz = \int_{0}^{\pi}\frac{e^{2i\epsilon e^{2i\theta}}}{\epsilon^2e^{4i\theta}} \epsilon e^{2i\theta}d\theta=2i\int_{0}^{\pi}\frac{e^{2i\epsilon e^{2i\theta}}}{\epsilon e^{2i\theta}}d\theta$$. But from here I cant send $\epsilon \to 0$, what mistake did I make, or is this approach completely wrong?

Answer 1

You don't need any $\epsilon$, $$\int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2}dx = \int_C \frac{\sin^2(z)}{z^2}dz = \int_C \frac{e^{2i z}-1}{-4z^2}dz+ \int_C \frac{e^{-2i z}-1}{-4z^2}dz$$ where $C$ is a piecewise linear curve $-\infty\to -1\to i \to 1\to +\infty$.

Those two integrals are found easily from the residue theorem adding to $C$ the 3 edges of an infinite square in the upper or lower half-plane.