Improper integral with sinus

Prove that:

$$\int_0^\infty \frac{\sin^2 x}{x^2}\, {\rm d}x = \frac{\pi}{2}$$

Solution

We present $3$ solutions. One is using contour integration and the other two are using Laplace Transformation.

1st solution:

By making use of the identity $\displaystyle \sin^2 x = \frac{1}{2} (1-\cos 2x)$ we have successively:

$$\begin{aligned}
\int_{0}^{\infty}\frac{\sin^2 x}{x^2}\, {\rm d}x &=\int_{0}^{\infty}\frac{\frac{1}{2}\left ( 1-\cos 2x \right )}{x^2}\, {\rm d}x \\
&= \int_{0}^{\infty}\frac{1-\cos y}{y^2}\, {\rm d}y\\
&= \cancelto{0}{\left [ - \frac{1-\cos y}{y} \right ]_0^\infty} + \int_{0}^{\infty} \frac{\sin y}{y}\, {\rm d}y
\end{aligned}$$

Now consider the function $\displaystyle f(z)=\frac{e^{iz}}{z}$ . Clearly it has a simple pole at $z=0$. We are integrating it at the following contour $\gamma$.

Since the pole is not included within the contour then by Cauchy Residue Theorem we have that $\displaystyle \oint_{\gamma} f(z)\, {\rm d}z=0$. Splitting the contour apart to the components it is consisted of we have:

$$\oint_{\gamma}f(z)\, {\rm d}z = \int_{-R}^{\epsilon}f(x)\, {\rm d}x + \oint_{\gamma_\epsilon}f(z)\, {\rm d}z + \int_{\epsilon}^{R}f(x)\, {\rm d}x + \oint_{\Gamma_R} f(z)\, {\rm d}z$$

Now, sending $\epsilon \rightarrow 0$ the contribution of the small circle is $-i \pi$, since:

$$\lim_{\epsilon \to 0} \int_{\gamma_{\epsilon}}^{} \! \frac{e^{iz}}{z} \, {\rm d}z =-i\lim_{\epsilon \to 0}\int_{-\pi}^0 \! e^{i\epsilon \cos(t)+\epsilon \sin(t)} \, {\rm d}t =-i\int_{-\pi}^0 \! 1 \, {\rm d}t=- i \pi$$

while sending $R\rightarrow +\infty$ the contribution of the large circle is $0$ , since:

$$\lim_{R\rightarrow +\infty}\left | \int_{\Gamma_R}\frac{e^{iz}}{z}\, {\rm d}z \right | \leq \lim_{R\rightarrow +\infty}\int_{0}^{\pi}\frac{{\rm d}t}{e^{R\sin t}}= \int_{0}^{\pi}0\, {\rm d}t =0 $$

Hence:

$$\int_{-\infty}^{0}\frac{e^{iz}}{z}\, {\rm d}z + \int_{0}^{\infty}\frac{e^{iz}}{z}\, {\rm d}z= i \pi \Leftrightarrow \int_{-\infty}^{\infty}\frac{e^{iz}}{z}\, {\rm d}z = i \pi$$

Taking imaginary parts yields the result. The proof also reveals that $\displaystyle \mathcal{P}\left ( \int_{-\infty}^{\infty} \frac{\cos x}{x}\, {\rm d}x \right )=0$.

2nd solution

$$\begin{aligned}
\int_{0}^{\infty}\frac{\sin x}{x}\, {\rm d}x &=\int_{0}^{\infty}\sin x \left( \int_{0}^{\infty}e^{-xt}\, {\rm d}t\right)\; {\rm d}x \\
&=\int_{0}^{\infty}\left ( \int_{0}^{\infty}\sin x e^{-xt}\, {\rm d}x \right )\, {\rm d}t \\
&=\int_{0}^{\infty}\frac{{\rm d}t}{t^2+1} = \frac{\pi}{2}
\end{aligned}$$

3rd solution (again Laplace but a little weird)

$$\begin{aligned}\int_{0}^{\infty}\dfrac{\sin^2\,x}{x^2}\,\mathrm{d}x&=\int_{0}^{\infty}\dfrac{\sin\,x}{x}\,\mathrm{d}x\\&=\int_{0}^{+\infty}e^{-x}\,\dfrac{e^{x}\,\sin\,x}{x}\,\mathrm{d}x\\&=\mathcal{L}\,\left(\dfrac{e^{x}\,\sin\,x}{x}\right)(1)\\&=\int_{1}^{+\infty}\mathcal{L}\,\left(e^{x}\,\sin\,x\right)(s)\,\mathrm{d}s\\&=\int_{1}^{\infty}\dfrac{1}{1+(s-1)^2}\,\mathrm{d}s\\&=\left[\arctan\,(s-1)\right]_{1}^{\infty}\\&=\dfrac{\pi}{2}\end{aligned}$$