How do I show that $\int_{-\infty}^\infty \frac{ \sin x \sin nx}{x^2} \ dx = \pi$? Quick question. Could somebody please explain to me why it is that
$$\int_{-\infty}^\infty \frac{ \sin x \sin nx}{x^2} \ dx = \pi$$
for every positive integer $n$? This integral showed up when I was computing a certain normalization constant. I was planning on just labeling it $I_n$ and moving on with my life but then Wolfram Alpha informed me it always equals $\pi$. Thanks!
 A: Another way to see why it should be so is to go to the frequency domain. Let $f_a(x)=\frac{\sin ax}x$ and $a\ge b>0\,$. The Fourier transform of $f_a(x)$ is a step:
$$
F[f_a](\xi)=\sqrt{\frac\pi2}\theta(a-|x|),
$$
there $\theta$ is the Heaviside step function.
By the properties of Fourier transform we have
$$
\int_{-\infty}^{\infty}f_a(x)f_b(x)\,dx=F[f_af_b](0)=
F[f_a]*F[f_b](0)=
$$
$$\int_{-\infty}^{\infty}F[f_a](\xi)F[f_b](-\xi)\,d\xi=
\frac\pi2 \int_{-b}^b d\xi=\pi b.
$$
A: Another approach is integration by parts:
$$
\begin{align}
\int_{-\infty}^\infty \frac{ \sin x \sin nx}{x^2} \mathrm dx
&=
\int_{-\infty}^\infty\frac{\cos x\sin nx+n\sin x\cos nx}x\mathrm dx
\\
&=
\frac12\int_{-\infty}^\infty\frac{\sin(n+1)x+\sin(n-1)x+n(\sin(n+1)x-\sin(n-1)x)}x\mathrm dx
\\
&=
\frac12(1+1+n-n)\int_{-\infty}^\infty\frac{\sin x}x\mathrm dx
\\
&=
\pi\;.
\end{align}
$$
For $n=1$, the terms with $\sin(n-1)x$ don't occur, but the result is the same.
A: Let $n\ge 1$. You can integrate $$f(z) = \frac{\sin(z)e^{inz}}{z^2}$$ around a big half-disc $U_R$ in the upper half-plane. The integral over the circle-part will go to $0$ for $R\to\infty$ (that's where $n\ge 1$ is needed). Therefore
$$\int_{-\infty}^\infty \frac{\sin(x)\sin(nx)}{x^2}\, dx = \lim_{R\to \infty} \mathrm{Im}\left[\oint_{\partial U_R} f(z) \, dz\right] = \mathrm{Im}\left[\pi i \;\mathrm{Res}_{z=0}(f(z))\right] = \pi$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{-\infty}^{\infty}{\sin\pars{x}\sin\pars{nx} \over x^{2}}\,\dd x = \pi
     :\ {\large ?}.\qquad n \in {\mathbb N}_{>0}}$

With the identity
  $\ds{{\sin{x} \over x} = \half\int_{-1}^{1}\expo{\pm\ic kx}\,\dd k}$:
  \begin{align}&\color{#c00000}{%
\int_{-\infty}^{\infty}{\sin\pars{x}\sin\pars{nx} \over x^{2}}\,\dd x}
=\int_{-\infty}^{\infty}\pars{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}
\pars{n\,\half\int_{-1}^{1}\expo{-\ic qnx}\,\dd q}\,\dd x
\\[3mm]&={1 \over 4}\,n\int_{-1}^{1}\int_{-1}^{1}\
\underbrace{\int_{-\infty}^{\infty}\expo{\ic\pars{k - nq}x}\,\dd x}
_{\ds{=\ 2\pi\,\delta\pars{k - nq}}}\ \,\dd k\,\dd q
\end{align}
  where $\ds{\delta\pars{x}}$ is the
  Dirac Delta Function.

\begin{align}&\color{#c00000}{%
\int_{-\infty}^{\infty}{\sin\pars{x}\sin\pars{nx} \over x^{2}}\,\dd x}
=\half\,n\pi\int_{-1}^{1}\Theta\pars{1 - n\verts{q}}\,\dd q
\end{align}
$\ds{\Theta\pars{x}}$ is the
Heaviside Step Function.

\begin{align}&\color{#c00000}{%
\int_{-\infty}^{\infty}{\sin\pars{x}\sin\pars{nx} \over x^{2}}\,\dd x}
=\half\,n\pi\int_{-1}^{1}\Theta\pars{{1 \over n} - \verts{q}}\,\dd q
=\half\,n\pi\int_{-1/n}^{1/n}\dd q=\half\,n\pi\pars{2 \over n}
\end{align}

$$
\color{#00f}{\large%
\int_{-\infty}^{\infty}{\sin\pars{x}\sin\pars{nx} \over x^{2}}\,\dd x
= \pi}\,\qquad\qquad n = 1,2,3,\ldots
$$
A: It depends how rigorous you want to be. For $n=1$ this is a classic integral, that I'll assume you have seen before/can easily find. For $n>1$ we have the following generalization if we let $a>b\geqslant 0$
$$\begin{aligned}2\int_0^{\infty}\frac{\sin(ax)\sin(bx)}{x^2}\;dx &=\int_0^{\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{x^2}\;dx \\ &= \int_0^{\infty}\int_{a-b}^{a+b}\frac{\sin(xy)}{x}\;dy \;dx\\ &=\int_{a-b}^{a+b}\int_0^{\infty}\frac{\sin(xy)}{x}\;dx \;dy \\ &=\int_{a-b}^{a+b}\frac{\pi}{2}\;dy \\ &=\pi b\end{aligned}$$
A: Consider, 
$$f(z) = \Im \frac{e^{iz[n+1]}}{z^2}$$
Consider a semi circle, with a small empty semi circle.

The residue is $0$
$$\oint_{C} f(z) dz = 0 = \int_{BDEFG} f(z) dz + \int_{-R}^{-\epsilon} f - (i)\int_{0}^{\pi} \frac{e^{i\epsilon e^{i\theta}}}{\epsilon e^{i\theta}} + \int_{\epsilon}^{R} f$$
$$\lim_{R \to \infty, \epsilon \to 0} \oint_{C} f(z) dz= 0 + \int_{-\infty}^{0} f(x) dx - (i)\int_{0}^{\pi} \lim_{\epsilon \to 0} \frac{e^{i\epsilon e^{i\theta}}}{\epsilon e^{i\theta}} + \int_{0}^{\infty} f(x) dx  \space \space \space \space \space \space \space \space \space \space \space \space \space (1)   $$    
$$\int_{-\infty}^{\infty} f(x) dx - (i)\int_{0}^{\pi} \lim_{\epsilon \to 0} \frac{e^{i\epsilon e^{i\theta}}}{\epsilon e^{i\theta}} = 0 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space  (2)$$
$$\int_{-\infty}^{\infty} f(x) dx = \Im \{i\pi\} = \pi$$
We took the imaginary part because we took the imaginary part of $e^{iz}$ to get $\sin(z)$ in the first place.
In $(1)$ $\displaystyle \int_{BDEFG} f(z) dz$ vanishes. Below is how: Read Jordan's Lemma first.
We see that:
$$f(z) = e^{iz[n+1]}g(z) \space \text{where} \space g(z) = \frac{1}{z^2}$$
$$M_R = \max|g(Re^{i\theta})| = \frac{1}{\left| Re^{i\theta} \right |^2} = \frac{1}{R^2}$$
Jordan's lemma states:
$$\left| \int_{BDEFG} f(z) dz \right| \le \frac{\pi}{n+1}M_R = \frac{\pi}{R^2(n+1)}$$
$$\lim_{R \to \infty} \frac{\pi}{R^2(n+1)} = 0$$ 
Therefore, the integral on the top contour vanished (becomes 0).
In $(2)$ we take the limit inside the integral due to the dominated convergence theorem. Find $g(z)$ such that, $|f(z)| \le g(z)$
