Having played around with Wolfram Alpha, I find the following improper integral:

$$\int_0^{\infty} \frac{(\sin x)^2}{x^2-\pi ^2} dx=-\frac1{2\pi}\tag{1} $$

(See also the Wolfram Alpha snapshot below.)

But I don't know how to prove this result. Without the singularity at $x=\pi$, there have been several questions about evaluating $\int_0^\infty\frac{\sin x}{x}\,dx$, which seems not to be very related to the one here. See for instance Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?

How can I prove (1)?

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    $\begingroup$ Cauchy and Jordan. $\endgroup$ – Jon Apr 4 '18 at 13:29
  • $\begingroup$ I checked it with WA. Once I got 0. Another time I got $-\frac{1}{2\pi}$. Really weird. $\endgroup$ – trancelocation Apr 4 '18 at 18:13

Actually $$ \mathcal{J}=\int_{0}^{+\infty}\frac{\sin^2 x}{x^2-\pi^2} = \color{red}{0}. $$ Indeed by parity $$ \mathcal{J}=\frac{1}{4}\int_{-\infty}^{+\infty}\frac{1-\cos(2x)}{x^2-\pi^2}\,dx =\frac{1}{4}\text{Re}\int_{-\infty}^{+\infty}\frac{1-e^{2ix}}{x^2-\pi^2}\,dx$$ and the meromorphic function $\frac{1-e^{2ix}}{x^2-\pi^2}$ fulfills the ML lemma and it is actually holomorphic.
By considering a semicircle contour in the upper half-plane, centered at the origin, having radius $R\to +\infty$ and with two small bulges (with radius $\varepsilon\to 0^+$) around $x=-\pi$ and $x=\pi$, it turns out that the original integral equals one fourth of the real part of the residue of some function... at no point!

Non-believers may try the Mathematica $\text{NIntegrate}$ command, or the alternative, more real-analytic approach

$$ \sum_{k\geq 0}\frac{1}{(x+k\pi)^2-\pi^2} = \frac{\pi-2x}{\pi x (\pi-x)}\quad\Longrightarrow\quad \mathcal{J}=\int_{0}^{\pi}\underbrace{\frac{\pi-2x}{2\pi x(\pi-x)}\sin^2(x)}_{\text{odd with respect to }x=\frac{\pi}{2}}\,dx = 0.$$

Curious bug of WA, the lesson probably is do not trust machines too much.

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    $\begingroup$ For the sake of completeness, Maple yields the right answer. $\endgroup$ – Jon Apr 4 '18 at 13:52
  • $\begingroup$ I have a question , would it be useful to use the taylor series expansion of $\sin(x)$ centered at $\pi$? $\endgroup$ – The Integrator Apr 4 '18 at 16:10
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    $\begingroup$ @pranavB23: no. It is not termwise-integrable over $\mathbb{R}$ or $\mathbb{R}^+$, and the situation is the same if you plug in the factor $\frac{1}{\pi^2-x^2}$ (which has simple poles at $x=\pm\pi$, by the way). $\endgroup$ – Jack D'Aurizio Apr 4 '18 at 16:12
  • $\begingroup$ @JackD'Aurizio oh ok , i get it now . Thank you :) $\endgroup$ – The Integrator Apr 4 '18 at 16:31
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    $\begingroup$ @Jon $\texttt{Mathematica}$ (in a MacBook Pro) yields the right answer too. $\endgroup$ – Felix Marin Apr 4 '18 at 21:06

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