Compute integral with residue theorem

Question

I have the following integral $$\int_0^{\infty} \frac{\sin^2x}{x^2}\mathrm{d}x$$ And a hint: integrate $\frac{e^{2iz}-1-2iz}{z^2}$ within a semi circle. But this function residue zero (what I understand as the coefficient of 1/z) what am I missing?

Answer 1

Brief solution. Consider the integral of the entire function $$f(z)=\frac{e^{2iz}-1-2iz}{z^2}\ \hbox{for $z\ne0$}\ ,\quad f(0)=-2$$ around a contour consisting of $C_1$, the line from $-R$ to $R$, and $C_2$, the upper semicircle from $R$ to $-R$.

For the reasons given in your question, $$\int_{C_1\cup C_2} f(z)\,dz=0\ .$$ Now $$\int_{C_2}f(z)\,dz=\int_{C_2}\frac{e^{2iz}-1}{z^2}\,dz-2i\int_{C_2}\frac{dz}z\ .$$ The first integral goes to zero as $R\to\infty$ and the second is evaluated by parametrisation. Also $$\int_{C_1} f(z)\,dz$$ can be evaluated in terms of the integral you want, and this solves the problem.

I'll leave you to work out the details.

Answer 2

Let,

$$I(a)=\int_{-\infty}^{\infty}\frac{\sin^2(ax)dx}{x^2}$$

$$\frac{dI}{da}=\int_{-\infty}^{\infty}\frac{\sin(2ax)dx}{x}=\pi$$

$$ I(a)=\pi a+ C$$

$$I(a)=\pi a$$ because $I(0)=0$.Hence for your particular case the answer is $ \pi /2$