# 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:

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

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

• For the sake of completeness, Maple yields the right answer. – Jon Apr 4 '18 at 13:52
• I have a question , would it be useful to use the taylor series expansion of $\sin(x)$ centered at $\pi$? – The Integrator Apr 4 '18 at 16:10
• @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). – Jack D'Aurizio Apr 4 '18 at 16:12
• @JackD'Aurizio oh ok , i get it now . Thank you :) – The Integrator Apr 4 '18 at 16:31
• @Jon $\texttt{Mathematica 10.0.0.0}$ (in a MacBook Pro) yields the right answer too. – Felix Marin Apr 4 '18 at 21:06