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.

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