Integral $\int_{-\infty}^{\infty} \frac{\sin^2x}{x^2} e^{ix}dx$ 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.
 A: 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 
$$\int_{-\infty}^{\infty} \frac{\sin^2x\cos x}{x^2} {\rm d}x=\pi\Big(1-\frac12\Big)=\frac{\pi}{2}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
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 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
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 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\int_{-\infty}^{\infty}\frac{\sin^{2}\pars{x}}{x^{2}}\,\expo{\ic x}\,\dd x:
     \ {\large ?}}$.
\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
  $\ds{\int_{-1}^{1}\delta\pars{1 + k - q}\,\dd q
=\left\{\begin{array}{lcl}
1 & \mbox{if} & -1 < 1 + k < 1
\\[2mm]
0 && \mbox{otherwise}
\end{array}\right.}$

A: 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,
$$
\begin{align}
&\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,e^{iax}\,\mathrm{d}x\\
&=\int_{-\infty}^\infty\frac{\sin^2(x)\cos(ax)}{x^2}\,\mathrm{d}x\\
&=\int_{-\infty}^\infty\frac{2\cos(ax)-\cos((a+2)x)-\cos((a-2)x)}{4x^2}\,\mathrm{d}x\\
&=\int_{-\infty}^\infty\frac{(a+2)\sin((a+2)x)+(a-2)\sin((a-2)x)-2a\sin(ax)}{4x}\,\mathrm{d}x\\[2pt]
&=\frac{|a+2|+|a-2|-2|a|}4\,\pi\\[6pt]
&=\left\{\begin{array}{}0&\text{if }|a|\ge2\\
\dfrac{2-|a|}2\,\pi&\text{if }|a|\lt2
\end{array}\right.
\end{align}
$$
A: $$\int_{-\infty}^{\infty} \frac{\sin^2x\cos x}{x^2} {\rm d}x=\frac14\int_{-\infty}^{\infty} \frac{\cos x-\cos3x}{x^2} {\rm d}x=\frac14\pi(3-1)=\frac{\pi}2$$
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.
A: 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$.
A: 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.
