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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

6 Answers 6

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
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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$$f(z) = \Im \frac{e^{iz[n+1]}}{z^2}$$

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, $|f(z)| \le g(z)$

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$$\Im e^{i(n+1)z}\neq\sin{z}\sin(nz)$$ –  M.N.C.E. Jan 4 at 15:21
@M.N.C.E., how? $$\sin(z) = \Im e^{iz}$$ then $$\sin(nz) = \Im e^{niz}$$ $$\sin(nz)\sin(z) = \Im e^{iz(n+1)}$$ –  Amad27 Jan 4 at 16:09

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