Integral $\int_0^{\infty} \frac{\sin^3{x}}{x} \; dx$ I need help with this integral, please: $$\int_0^{\infty} \frac{\sin^3{x}}{x} \; dx$$
I know the Dirichlet integral $\int_0^{\infty} \frac{\sin{x}}{x} \; dx=\frac{\pi}{2}$ but what about with the cube of $\sin{x}$ in the numerator?  I tried using Feynman's method with $e^{ax}$ at $a=0$ but I couldn't find anything helpful.
 A: notice that:
$$\sin^3(x)=\frac{3\sin(x)-\sin(3x)}{4}$$
now if we have:
$$I=\int_0^\infty\frac{\sin^3(x)}{x}dx=\frac{1}{4}\int_0^\infty\frac{3\sin(x)-\sin(3x)}{x}dx=J+K$$
$$J=\frac 34\int_0^\infty\frac{\sin(x)}{x}dx=\frac{3\pi}{8}$$
$$K=-\frac 14\int_0^\infty\frac{\sin(3x)}{x}dx$$
$u=3x$, $du=3dx$ and so: $dx=\frac {du}3$ which gives:
$$K=-\frac 14\int_0^\infty\frac{\sin(u)}{u/3}\frac{du}3=-\frac 14\int_0^\infty\frac{\sin(u)}{u}du=-\frac{\pi}{8}$$
and so:
$$I=\frac{3\pi}{8}-\frac{\pi}{8}=\frac{\pi}{4}$$
A: Note that you can express $\sin^3(x)$ in terms of linear sine terms as follows:
$$\sin^3(x)={\left(\frac{e^{ix}-e^{-ix}}{2i}\right)}^3=-\frac{1}{4} \left(\frac{e^{3ix}-e^{-3ix}-3e^{ix}+3e^{-ix}}{2i}\right)=-\frac{1}{4} \left(\sin{(3x)}-3\sin{x}\right)$$
Thus, the integral is  \begin{align*} \int_0^{\infty} \frac{\sin^3{x}}{x} \; \mathrm{d}x&=\frac{3}{4} \int_0^{\infty} \frac{\sin{x}}{x} \; \mathrm{d}x - \frac{1}{4} \underbrace{\int_0^{\infty} \frac{\sin{(3x)}}{x} \; \mathrm{d}x}_{3x \to x}\\ 
&=\frac{1}{2} \int_0^{\infty} \frac{\sin x}{x} \; \mathrm{d}x \\
&= \boxed{\frac{\pi}{4}}
\end{align*}
Where the final integral is recognized as the Dirichlet integral.
