How can I prove the Wolfram Alpha result $\int_0^{\infty} \frac{(\sin x)^2}{x^2-\pi ^2} dx=-\frac1{2\pi}$? 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)?




 A: 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.
A: I will present a more elementary approach. First, note that if we expand the numerator at the singularity, first term in the Taylor series of the numerator $\sin^2{x}$ is $(x-\pi)^2$, and the denominator is $(x-\pi)(x+ \pi)$, so the singularity does not contribute to the integral (we can just remove the point $x = \pi$ from the region of integration), and we can treat it with elementary  methods.
$$\int_0^\infty \frac{\sin^2{x}}{(x^2-\pi^2)} dx =\frac{1}{2} \int_{-\infty}^\infty \frac{\sin^2{x}}{(x^2-\pi^2)} dx \text{    (since even function)}$$
$$= \frac{1}{2} \int_{-\infty}^\infty \frac{\sin^2{x}}{(x-\pi)(x + \pi)} dx = \frac{1}{4\pi}\int_{-\infty}^\infty {\sin^2{x}}\left(\frac{1}{x-\pi} - \frac{1}{x+\pi}\right) dx  $$
$$= \frac{1}{4\pi}\int_{-\infty}^\infty \frac{{\sin^2{x}}}{x-\pi}  dx- \frac{1}{4\pi}\int_{-\infty}^\infty \frac{{\sin^2{x}}}{x+\pi}dx$$
Now substitute $(x - \pi) = t$ and $(x + \pi) = s$ in the two integrals.
Then the expression becomes $$\frac{1}{4\pi}\int_{-\infty}^\infty \frac{{\sin^2{(t + \pi)}}}{t} dt - \frac{1}{4\pi}\int_{-\infty}^\infty \frac{{\sin^2{(s-\pi)}}}{s} ds$$
$$= \frac{1}{4\pi}\int_{-\infty}^\infty \frac{{\sin^2{t}}}{t} dt - \frac{1}{4\pi}\int_{-\infty}^\infty \frac{{\sin^2{s}}}{s} ds = 0$$
