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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!

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Here's a fun "not a proof": Replace $n$ by a real parameter $t$, call the integral $f(t)$. For $t>0$ We have $f'(t) "=" \int t \frac{\sin(x) \cos(tx)}{x^2}$, which is $0$ since the integrand is odd (or would be $0$, if it converged and we could legally differentiate under the integral sign). The same argument "proves" the integral is constant if you replace $\infty$ by $1$, which is false. –  Kevin Costello Feb 7 '12 at 7:35
    
You can use Residue theorem –  user24608 Feb 8 '12 at 13:43

5 Answers 5

up vote 13 down vote accepted

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 $a>b\geqslant 0$

$$\begin{aligned}2\int_0^{\infty}\frac{\sin(ax)\sin(bx)}{x^2}\;dx &=\int_0^{\infty}\frac{\cos((a-b)x)-\cos((a+b)x)}{x^2}\;dx \\ &= \int_0^{\infty}\int_{a-b}^{a+b}\frac{\sin(xy)}{x}\;dy \;dx\\ &=\int_{a-b}^{a+b}\int_0^{\infty}\frac{\sin(xy)}{x}\;dx \;dy \\ &=\int_{a-b}^{a+b}\frac{\pi}{2}\;dy \\ &=\pi b\end{aligned}$$

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Thanks for the reply. Unfortunately, I can't recall seeing the $n=1$ case at any point and can't seem to find it anywhere. I like your derivation. It's not quite a proof, but still fairly satisfying. One of the more troublesome things to deal with might be that $\int_0^\infty \left| \frac{\sin xy}{x} \right| \ dx = \infty$ - no? –  Mike F Feb 7 '12 at 7:44
    
Could someone edit this by putting $\mathrm dx$ and $\mathrm dy$ in the appropriate places, please? –  John Bentin Feb 7 '12 at 7:45
    
I agree with Mike. The third line has to be detailled. It is not clear for me that you have to assume $a>b \geq 0$. –  user10676 Feb 7 '12 at 9:55
    
@Mike: You are correct, so that Fubini cannot be directly applied. That said if you write $x^{-2}$ as $\displaystyle \int_0^{\infty}te^{-tx}$ you can make everything rigorous with Tonelli's. –  Alex Youcis Feb 7 '12 at 22:27

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 $$ \int_{-\infty}^{\infty}f_a(x)f_b(x)\,dx=F[f_af_b](0)= F[f_a]*F[f_b](0)= $$ $$\int_{-\infty}^{\infty}F[f_a](\xi)F[f_b](-\xi)\,d\xi= \frac\pi2 \int_{-b}^b d\xi=\pi b. $$

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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.

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+1 for the $1+1+n-n$ step. –  Did May 6 '12 at 12:48
    
Thanks this is nice! Summary: integrate by parts (differentiating the top); make use of the identity $2 \sin A \cos B = \sin(A+B) + \sin (A-B)$; notice that $\int_{-\infty}^\infty \frac{g(ax)}{x} dx = \int_{-\infty}^\infty \frac{g(x)}{x} dx$ ($g$ suitable, $a>0$); know that $\int_{-\infty}^\infty \frac{\sin x}{x} dx = \pi$. –  Mike F May 9 '12 at 4:05
1  
I do find it a tad strange that $\int \frac{sin x}{x}$ keeps showing up in these derivations since that's only a "conditionally convergent" integral and I should think the original integral converges in the absolute sense - since it is bounded at $x=0$ and decays like $1/x^2$... –  Mike F May 9 '12 at 4:12
    
@Mike: That's an interesting observation. –  joriki May 9 '12 at 4:21

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

$$\int_{-\infty}^\infty \frac{\sin(x)\sin(nx)}{x^2}\, dx = \lim_{R\to \infty} \mathrm{Im}\left[\oint_{\partial U_R} f(z) \, dz\right] = \mathrm{Im}\left[\pi i \;\mathrm{Res}_{z=0}(f(z))\right] = \pi$$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{-\infty}^{\infty}{\sin\pars{x}\sin\pars{nx} \over x^{2}}\,\dd x = \pi :\ {\large ?}.\qquad n \in {\mathbb N}_{>0}}$

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}

$$ \color{#00f}{\large% \int_{-\infty}^{\infty}{\sin\pars{x}\sin\pars{nx} \over x^{2}}\,\dd x = \pi}\,\qquad\qquad n = 1,2,3,\ldots $$

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