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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?

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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.

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  • $\begingroup$ So would the contour just be the infinite square? $\endgroup$
    – Joey
    Dec 9, 2020 at 6:29
  • $\begingroup$ "the contour" ? the three edges of an infinite square, the 4th edge is $C$. $\endgroup$
    – reuns
    Dec 9, 2020 at 6:39
  • $\begingroup$ 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 $\endgroup$
    – Joey
    Dec 9, 2020 at 6:45
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    $\begingroup$ $C$ 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. $\endgroup$
    – reuns
    Dec 9, 2020 at 6:49

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