Definite Integral to Infinity of Cube of Sine x over x Cubed

Theorem

 * $\ds \int_0^\infty \frac {\sin^3 x} {x^3} \rd x = \frac {3 \pi} 8$

Proof
Let:


 * $\ds \map I \alpha = \int_0^\infty \frac {\map {\sin^3} {\alpha x} } {x^3} \rd x$

for positive real parameter $\alpha$.

We have:

We aim to evaluate explicitly:


 * $\ds \int_0^\infty \frac {\sin^3 x} {x^3} \rd x = \map I 1$

Differentiating with respect to $\alpha$ we have:

We therefore have:

By Fundamental Theorem of Calculus: Second Part, we also have:


 * $\ds \int_0^1 \map {I'} \alpha \rd \alpha = \map I 1 - \map I 0 = \map I 1$

giving:


 * $\ds \int_0^\infty \frac {\sin^3 x} {x^3} \rd x = \frac {3 \pi} 8$