Calculating $\int_0^\infty \frac {\sin^2x}{x^2}dx$ using the Residue Theorem. [duplicate]
Hint. First note that $$ \sin^2x=\frac{1}{2}\mathrm{Re}\, \big(1-\mathrm{e}^{2xi}\big), $$ and hence your integral equals $$ \frac{1}{4}\int_{-\infty}^{\infty}\frac{1-\mathrm{e}^{2xi}}{x^2}dx $$ Then define the curve $\gamma_{\varepsilon,R}$ to be the union of:
$\gamma_R(t)=R\mathrm{e}^{it}, \,\,t\in[0,\pi]$.
$\gamma_-(t)=t, \,\,t\in[-R,\varepsilon],$
$\gamma_\varepsilon(\pi-t),\,\,t\in[0,\pi],$
$\gamma_+(t)=t,\,\,t\in[\varepsilon,R]$.
Then use Residue Theorem (here the function DOES NOT have poles in the interior of $\gamma_{\varepsilon,R}$), and let $\varepsilon\to0$, $R\to\infty$.
ANSWER. $\dfrac{\pi}{2}$.
Hint: Write $\sin^2(x) = \frac{1}{2}Re(1 - e^{2ix})$.
Take the integration as $$\int \frac{1 - e^{2iz}}{z^2} dz$$
Take the contour s.t. $0 + 0i$ is out of your closed region.
Find residue and solve.
In this answer, it is shown using contour integration and the binomial theorem that $$ \int_0^\infty\left(\frac{\sin(z)}{z}\right)^n\mathrm{d}z =\frac{\pi}{2^n(n-1)!}\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\binom{n}{k}(n-2k)^{n-1} $$ Plugging in $n=2$ gives $\dfrac\pi2$.