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From answer of Calculating $\int_0^\infty \frac {\sin^2x}{x^2}dx$ using the Residue Theorem., I believe the non-zero part is coming from computing $$\displaystyle\int\frac{1-e^{2iz}}{z^2}dz$$ along $\gamma_\epsilon$, the small semi-circular contour around zero.

But since $(\sin z/z)$ has a removable singularity, it will be bounded near $0$, which means $(\sin z/z)^2$ bounded near $0$...Doesn't that mean by ML inequality the integral goes to $0$ as $\epsilon \to 0$?

Lalit Tolani
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SmoothKen
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  • But $\frac{1-e^{2iz}}{z^2}$ has a simple pole at $z=0$. – Sangchul Lee Sep 01 '21 at 16:37
  • @SangchulLee Yes. I agree. But why does my reasoning above not work? – SmoothKen Sep 01 '21 at 16:57
  • If you use the function $\frac{\sin^2 z}{z^2}$ instead, then its integral along the simicircular arc will not vanish as its radius goes to infinity, and it will be too wild to control. That is why we bother to use $\frac{1-e^{2iz}}{z^2}$ instead. – Sangchul Lee Sep 01 '21 at 17:32

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