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

$$\int_{-\infty}^{\infty} \frac{\sin^2x\cos x}{x^2} {\rm d}x=\frac14\int_{-\infty}^{\infty} \frac{\cos x-\cos3x}{x^2} {\rm d}x=\frac14\pi(3-1)=\frac{\pi}2$$ 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.

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$.

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 $$\int_{-\infty}^{\infty} \frac{\sin^2x\cos x}{x^2} {\rm d}x=\pi\Big(1-\frac12\Big)=\frac{\pi}{2}.$$