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How do I determine the value of this integral?

$$\int_{-\infty}^{\infty} \frac{\sin^2x}{x^2} e^{ix}dx$$

Plugging in Euler's identity gives

$$\int_{-\infty}^{\infty} \frac{i\sin^3x}{x^2}dx + \int_{-\infty}^{\infty} \frac{\sin^2x \cos x}{x^2}dx$$ and since $\dfrac{i\sin^3x}{x^2}$ is an odd function, all that is left is

$$\int_{-\infty}^{\infty} \frac{\sin^2x \cos x}{x^2}dx$$

at which point I am stuck.

I feel I am not even going the right direction, anybody willing to help?

Thanks in advance.

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  • $\begingroup$ Did you try the residue theorem? $\endgroup$ – Tryss Feb 20 '15 at 16:40
  • $\begingroup$ If you know about Fourier transforms, you can use that this integral is $F(-1)$, where $F(\xi)=\int_{-\infty}^\infty f(x) e^{-i\xi x}$ is the Fourier transform of $f(x)=\left( \frac{\sin x}{x} \right)^2$. The Fourier transform of $\frac{\sin x}{x}$ is known to be a rectangular pulse, and when you square it, you take the convolution on the transform side, so that $F(\xi)$ becomes a tent function. $\endgroup$ – Hans Lundmark Feb 22 '15 at 18:53
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$$\int_{-\infty}^{\infty} \frac{\sin^2x\cos x}{x^2} {\rm d}x=\frac14\int_{-\infty}^{\infty} \frac{\cos x-\cos3x}{x^2} {\rm d}x=\frac14\pi(3-1)=\frac{\pi}2$$ where I used: $$\int_{-\infty}^{\infty}\frac{\cos (ax)-\cos(bx)}{x^2}=\pi(b-a)\tag{$b,a>0$}$$ which is easily obtained using countour integration. Prooved here.

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For a probabilistic approach, integration by parts leads to: $$I=\int_{\mathbb{R}}\frac{\sin^2 x\cos x}{x^2}\,dx = \frac{2}{3}\int_{\mathbb{R}}\left(\frac{\sin x}{x}\right)^3\,dx \tag{1}$$ but since $\frac{\sin t}{t}$ is the CF of the uniform distribution over the interval $[-1,1]$, by assuming that $X_1,X_2,X_3$ are identically distributed and indipendent, $X_i$ is uniformly distributed over $[-1,1]$, $Z=X_1+X_2+X_3$ and $f_Z$ is the PDF of $Z$, we have: $$ I = \frac{4\pi}{3}\cdot f_Z(0) = \frac{4\pi}{3}\cdot\frac{3}{8} =\color{red}{\frac{\pi}{2}}.\tag{2}$$ With the same approach we can also compute, for any $n\geq 2$, $ \int_{\mathbb{R}}\left(\frac{\sin x}{x}\right)^n\,dx$.

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    $\begingroup$ what?what?what? Please expand, it's too much info quick overdose! $\endgroup$ – RE60K Feb 20 '15 at 17:15
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    $\begingroup$ @ADG: LOL. By the way, I am just using the following property of the characteristic function (en.wikipedia.org/wiki/…): $$\varphi_{X_1+X_2}(t) = \varphi_{X_1}(t)\cdot \varphi_{X_2}(t).$$ $\endgroup$ – Jack D'Aurizio Feb 20 '15 at 17:20
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    $\begingroup$ OK just turns out that is not my level so you know .... i made that comment... uh sorry? $\endgroup$ – RE60K Feb 20 '15 at 17:25
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Define $\displaystyle I(a,b)=\int_{-\infty}^{\infty} \frac{\sin^2ax\cos bx}{x^2} {\rm d}x \displaystyle$, and take Laplace transform of $I(a,b)$ with respect to $a$ and $b$, with Laplace domain variables $s$ and $t$, respectively: \begin{align} \mathcal{L}_{a\rightarrow s,b\rightarrow t}\{I(a,b)\}&=\int_{-\infty}^{\infty} \frac{2t}{(t^2+x^2)(s^3+4sx^2)} {\rm d}x\\ &=\frac{2\pi}{s^2(s+2t)} \end{align} Now taking inverse Laplace with respect to $t$ and then with respect to $s$ we obtain

\begin{align} \mathcal{L}^{-1}_{s\rightarrow a}\Big\{\mathcal{L}^{-1}_{t\rightarrow b}\{\frac{2\pi}{s^2(s+2t)}\}\Big\}&=\mathcal{L}^{-1}_{s\rightarrow a}\{\pi\frac{e^{-\frac{bs}{2}}}{s^2}\}\\ &=\pi\Big(a-\frac{b}{2}\Big)H(a-\frac{b}{2}), \end{align} where $H(x)$ is the Heaviside function. Therefore for $a>\frac{b}{2}$we have $$I(a,b)=\int_{-\infty}^{\infty} \frac{\sin^2ax\cos bx}{x^2} {\rm d}x=\pi\Big(a-\frac{b}{2}\Big)$$ and for $a\leq\frac{b}{2}$ $$I(a,b)=\int_{-\infty}^{\infty} \frac{\sin^2ax\cos bx}{x^2} {\rm d}x=0.$$ This hence implies that $$\int_{-\infty}^{\infty} \frac{\sin^2x\cos x}{x^2} {\rm d}x=\pi\Big(1-\frac12\Big)=\frac{\pi}{2}.$$

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{-\infty}^{\infty}\frac{\sin^{2}\pars{x}}{x^{2}}\,\expo{\ic x}\,\dd x: \ {\large ?}}$. \begin{align}&\color{#66f}{\large% \int_{-\infty}^{\infty}\frac{\sin^{2}\pars{x}}{x^{2}}\,\expo{\ic x}\,\dd x} =\int_{-\infty}^{\infty}\expo{\ic x}\,\ \overbrace{% \pars{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}}^{\dsc{\frac{\sin\pars{x}}{x}}} \ \overbrace{% \pars{\half\int_{-1}^{1}\expo{-\ic qx}\,\dd q}}^{\dsc{\frac{\sin\pars{x}}{x}}}\ \,\dd x \\[5mm]&=\frac{\pi}{2}\int_{-1}^{1}\int_{-1}^{1}\ \overbrace{% \int_{-\infty}^{\infty}\expo{\ic\pars{1 + k - q}x}\,\frac{\dd x}{2\pi}} ^{\dsc{\delta\pars{1 + k - q}}}\,\dd q\,\dd k =\frac{\pi}{2}\int_{-1}^{1}\int_{-1}^{1}\delta\pars{1 + k - q}\,\dd q\,\dd k \\[5mm]&=\left.\frac{\pi}{2}\int_{-1}^{1}\,\dd k\, \right\vert_{\, -1\ <\ 1 + k\ <\ 1} =\left.\frac{\pi}{2}\int_{-1}^{1}\,\dd k\,\right\vert_{\, -2\ <\ k\ <\ 0} =\frac{\pi}{2}\int_{-1}^{0}\,\dd k=\color{#66f}{\large\frac{\pi}{2}} \end{align}

Note that $\ds{\int_{-1}^{1}\delta\pars{1 + k - q}\,\dd q =\left\{\begin{array}{lcl} 1 & \mbox{if} & -1 < 1 + k < 1 \\[2mm] 0 && \mbox{otherwise} \end{array}\right.}$

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  • $\begingroup$ Not a mathematical solution. $\endgroup$ – Did Feb 23 '15 at 19:24
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We can get a bit more generality using the trigonometric identity $$ 4\sin^2(x)\cos(ax)=2\cos(ax)-\cos((a+2)x)-\cos((a-2)x) $$ Since the even part of $e^{iax}$ is $\cos(ax)$, we get, with an integration by parts, $$ \begin{align} &\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,e^{iax}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty\frac{\sin^2(x)\cos(ax)}{x^2}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty\frac{2\cos(ax)-\cos((a+2)x)-\cos((a-2)x)}{4x^2}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty\frac{(a+2)\sin((a+2)x)+(a-2)\sin((a-2)x)-2a\sin(ax)}{4x}\,\mathrm{d}x\\[2pt] &=\frac{|a+2|+|a-2|-2|a|}4\,\pi\\[6pt] &=\left\{\begin{array}{}0&\text{if }|a|\ge2\\ \dfrac{2-|a|}2\,\pi&\text{if }|a|\lt2 \end{array}\right. \end{align} $$

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The integral can be evaluate by converting it to a double integral first.

Let $$I = \int_{-\infty}^\infty \frac{\sin^2 x \cos x}{x^2} \, dx = 2 \int_0^\infty \frac{\sin^2 x \cos x}{x^2} \, dx.$$ Noting that $$\sin^2 x \cos x = \frac{1}{4} (\cos x - \cos 3x),$$ yields $$I = \frac{1}{2} \int_0^\infty \frac{\cos x - \cos 3x}{x^2} \, dx.$$

Now, by observing that $$\int_1^3 \sin (tx) \, dt = \frac{\cos x - \cos 3x}{x},$$ the integral for $I$ can be rewritten as $$I = \frac{1}{2} \int_0^\infty \int_1^3 \frac{\sin (xt)}{x} \, dt \, dx,$$ or $$I = \frac{1}{2} \int_1^3 \int_0^\infty \frac{\sin (xt)}{x} \, dx \, dt,$$ after the order of integration has been changed.

Enforcing a substitution of $x \mapsto x/t$ leads to $$I = \frac{1}{2} \int_1^3 \int_0^\infty \frac{\sin x}{x} \, dx \, dt.$$ Making use of the well-known result of $$\int_0^\infty \frac{\sin x}{x} \, dx = \frac{\pi}{2},$$ one has $$I = \frac{\pi}{4} \int_1^3 dt,$$ or $$\int_{-\infty}^\infty \frac{\sin^2 x \cos x}{x^2} \, dx = \frac{\pi}{2},$$ as expected.

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