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

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?

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.
• "the contour" ? the three edges of an infinite square, the 4th edge is $C$. Dec 9, 2020 at 6:39
• well the shape of $C$ was my question, but it would be a square with vertices at that goes to (-1,0), (-1,i),(1,i),(1,0) which goes to infinity or am I misunderstanding
• $C$ is a curve going from $-\infty$ to $\infty$ and passing above $0$. $\int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2}dx = \int_C \frac{\sin^2(z)}{z^2}dz$ by the Cauchy integral theorem. We can't close the contour because $\sin^2(z)$ grows very fast so we have to split the integral in $\int_C \frac{e^{2i z}-1}{-4z^2}dz+ \int_C \frac{e^{-2i z}-1}{-4z^2}dz$, then we can close the contour in each integral separately. Dec 9, 2020 at 6:49