I have seen in my complex analysis class that $lim_{\epsilon\to 0}\int\limits_{|x|\geq \epsilon}\frac{1-e^{ix}}{x^2}dx=\pi$.
From here, by taking the real part, we concluded that $\int\limits_{-\infty}^{\infty}\frac{1-cos(x)}{x^2}dx=\pi$.
I thought that now by taking to imaginary part, one should get $\int\limits_{-\infty}^{\infty}\frac{sin(x)}{x^2}dx=0$.
However, it does seem to me that it even converges, since near $0$ it looks like $1/x$ from both sides.
On the other hand, it is an odd function, so maybe it does exist and is $0$?
You can see the same argument in the following notes: Example 2
Thanks!
Answer
Through complex Analysis you may only show that
$$\color{red}{\text{PV}}\int_{-\infty}^{+\infty}\frac{\sin x}{x^2}\,dx = 0 \tag{1} $$
since we are not allowed to integrate through a pole, and $\int_{-\infty}^{+\infty}\frac{\sin x}{x^2}\,dx$ is not convergent in the usual sense (as an improper Riemann integral). On the other hand $(1)$ is trivial since $\frac{\sin x}{x^2}$ is an odd function.
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