how to calculate the integral of $\sin^2(x)/x^2$ [duplicate]

Solution 1:

An easy way:

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

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

$$\implies I(a)=\pi a+const$$

$$\implies I(a)=\pi a$$ because $I(0)=0$

Solution 2:

First we split $\sin^2(x)=\frac{(1-e^{2ix})+(1-e^{-2ix})}{4}$. To avoid the pole at $x=0$, drop the path of integration a bit below the real line (this function has no poles and it vanishes at infinity, so this is okay).

Next, let $\gamma^+$ be the path below the real axis, then circling back in a semi-circular path counterclockwise around the upper half-plane; and let $\gamma^-$ be the path below the real axis, then circling back in a semi-circular path clockwise around the lower half-plane.

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Note that $\gamma^+$ circles the pole at $x=0$ of $\frac{(1-e^{2ix})}{4x^2}$ and $\gamma^-$ misses the pole at $x=0$ of $\frac{(1-e^{-2ix})}{4x^2}$.

Therefore, $$ \begin{align} \int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\mathrm{d}x &=\int_{-\infty-i}^{\infty-i}\frac{1-\cos(2x)}{2x^2}\mathrm{d}x\\ &=\int_{-\infty-i}^{\infty-i}\frac{(1-e^{2ix})+(1-e^{-2ix})}{4x^2}\mathrm{d}x\\ &=\color{green}{\int_{\gamma^+}\frac{(1-e^{2ix})}{4x^2}\mathrm{d}x}+\color{red}{\int_{\gamma^-}\frac{(1-e^{-2ix})}{4x^2}\mathrm{d}x}\\ &=\color{green}{2\pi i\frac{-2i}{4}}+\color{red}{0}\\ &=\pi \end{align} $$

Solution 3:

Use Parseval theorem $\int_{-\infty}^{\infty}dx |f(x)|^{2}= \int_{-\infty}^{\infty}du|F(u)|^{2} $

the Fourier inverse transform of $ \frac{sin(x)}{x} $ is an step function (window function )