<s>[INST] An integral about trigonometric function. Recently I met an integral which is $\int_0^\infty \left(\frac{\sin x}{x}\right)^3 \; dx$.  I get the result is $3\pi/8$ by using Mathematica,
but I cannot derive it independently.
So I hope someone can help me.
It is my first time to ask questions on math.stackexchange.
Please tell me if I have done something wrong, thanks.
At last, thanks all the people for your precious time.
 [/INST] Integrating by parts:
$$
   \int_0^\infty \frac{\sin^3(x)}{x^3} \mathrm{d}x = \int_0^\infty \sin^3(x) \mathrm{d} \left(-\frac{1}{2 x^2} \right) = \frac{3}{2} \int_0^\infty \cos(x) \frac{\sin^2(x)}{x^2} \mathrm{d}x = \ldots
$$
Integrating by parts again:
$$
     \ldots = \frac{3}{2} \int_0^\infty \frac{2 \cos^2(x) \sin(x) - \sin^2(x)}{x} \mathrm{d}x  = \ldots
$$
Using
$$
   2 \cos^2(x) \sin(x) - \sin^2(x) = \frac{1}{4} \left(3 \sin(3x)-\sin(x)\right)
$$
we continue:
$$
   \ldots = \frac{3}{8} \left( \underbrace{\int_0^\infty \frac{3 \sin(3x)}{x} \mathrm{d}x }_{3 \frac{\pi}{2}}- \underbrace{\int_0^\infty \frac{\sin(x)}{x} \mathrm{d}x}_{\frac{\pi}{2}}  \right) = \frac{3}{8} \pi
$$
where we used the value of the Dirichlet integral $\int_0^\infty \sin(x) \frac{\mathrm{d}x}{x} = \frac{\pi}{2}$ and a change of variable 
$$
   \int_0^\infty \frac{\sin(3x)}{x} \mathrm{d}x \stackrel{3x = y}{=} \int_0^\infty \frac{\sin(y)}{y} \mathrm{d}y = \frac{\pi}{2}
$$
</s>