Integral exercise asked by Diego Alvariz in the calculus form group

Exercise:

Integrate, $\int_{0}^{\infty }{\frac{{{\sin }^{3}}x}{x}dx}$

Solution: ${{\sin }^{3}}x=\sin x{{\sin }^{2}}x=\sin x\left( 1-{{\cos }^{2}}x \right)=\sin x-\sin x{{\cos }^{2}}x$

$=\sin x-\frac{1}{2}\sin 2x\cos x=\sin x-\frac{1}{4}\left( \sin \left( 2x-x \right)+\sin \left( 2x+x \right) \right)=\sin x-\frac{1}{4}\left( \sin x+\sin 3x \right)$

$=\frac{4\sin x-\sin x+\sin 3x}{4}=\frac{1}{4}\left( 3\sin x+\sin 3x \right)$

So $\int_{0}^{\infty }{\frac{{{\sin }^{3}}x}{x}dx}=\frac{3}{4}\int_{0}^{\infty }{\frac{\sin x}{x}dx+\frac{1}{4}\int_{0}^{\infty }{\frac{\sin 3x}{x}dx}}$

Let $u=3x\Leftrightarrow du=3dx$
So $\int_{0}^{\infty }{\frac{\sin 3x}{x}dx=\frac{1}{3}\int_{0}^{\infty }{\frac{\sin u}{\frac{u}{3}}du=\int_{0}^{\infty }{\frac{\sin u}{u}du}}}$

Now the aim to prove $\int_{0}^{\infty }{\frac{\sin x}{x}dx}=\frac{\pi }{2}$

To compute , $\int_{0}^{\infty }{\frac{\sin x}{x}dx}$ we will define a complex function $f\left( z \right)=\frac{{{e}^{iz}}}{z}$

This function is analytic on $\mathbb{C}$ , with its only pole being simple pole at origin .

Let $r>>0$ be a large positive real number and let $\varepsilon >0$ be a small positive real number

Define a contour $\gamma $ consisting of 4 pieces :

                         $\gamma =\left[ -r,-\varepsilon  \right]\cup {{\gamma }_{\varepsilon }}\cup \left[ \varepsilon ,r \right]\cup {{\gamma }_{r}}$

Where ${{\gamma }_{\varepsilon }}$ is the upper half –circle of radius $\varepsilon $ ( clockwise ) and

${{\gamma }_{r}}$ is the upper half –circle of radius $r$ , (contour clockwise ) as $\gamma $ is closed curve so

By Cauchy theorem we get $\int_{\gamma }{f\left( z \right)dz=0}$ and note that on ${{\gamma }_{\varepsilon _{{}}^{{}}}}$ we have $f\left( z \right)=\frac{1}{z}$

Since ${{\gamma }_{\varepsilon }}$ is the half-circle moves in clockwise we get

$\underset{\varepsilon \to 0}{\mathop{\lim }}\,\int_{{{\gamma }_{\varepsilon }}}{\frac{{{e}^{iz}}}{z}dz}=-i\underset{\varepsilon \to 0}{\mathop{\lim }}\,\int_{-\pi }^{0}{{{e}^{i\varepsilon \left( \cos t+\sin t \right)}}dt}=-i\int_{\pi }^{0}{dt}=-i\pi $

$\underset{r\to \infty }{\mathop{\lim }}\,\left| \int_{{{\gamma }_{r}}}{\frac{{{e}^{iz}}}{z}dz} \right|\le \underset{r\to \infty }{\mathop{\lim }}\,\int_{0}^{\pi }{\frac{1}{{{e}^{r\sin t}}}dt=\int_{0}^{\pi }{0dt=0}}$

Hence , $\int\limits_{\gamma }{f\left( z \right)dz}=\int_{-r}^{-\varepsilon }{f\left( z \right)dz}+\int_{{{\gamma }_{\varepsilon }}}{f\left( z \right)dz+\int_{\varepsilon }^{r}{f\left( z \right)dz+\int_{{{\gamma }_{r}}}{f\left( z \right)dz}=0}}$

$\Leftrightarrow \int_{-r}^{-\varepsilon }{f\left( z \right)dz}-i\pi +\int_{\varepsilon }^{r}{f\left( z \right)dz}=0\Leftrightarrow \int_{-r}^{-\varepsilon }{f\left( z \right)dz}+\int_{\varepsilon }^{r}{f\left( z \right)dz=i\pi }$

But $\int_{-\infty }^{\infty }{\frac{\sin x}{x}dx}=\operatorname{Im}\left( \int_{-\infty }^{\infty }{\frac{{{e}^{iz}}}{z}dz} \right)=\operatorname{Im}\left( i\pi  \right)=\pi $

Also $f\left( -x \right)=\frac{\sin \left( -x \right)}{-x}=\frac{\sin x}{x}=f\left( x \right)$ is even function

Thus $\int_{-\infty }^{\infty }{\frac{\sin x}{x}dx}=2\int_{0}^{\infty }{\frac{\sin x}{x}dx\Leftrightarrow }\int_{0}^{\infty }{\frac{\sin x}{x}=\frac{\pi }{2}}$

Therefore , $\int_{0}^{\infty }{\frac{{{\sin }^{3}}x}{x}dx}=\frac{3}{4}\left( \frac{\pi }{2} \right)-\frac{\pi }{8}=\frac{2\pi }{8}=\frac{\pi }{4}$