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

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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:

\(\ds \map I 0\) \(=\) \(\ds \int_0^\infty \frac {\map {\sin^3} {0 x} } {x^3} \rd x\)
\(\ds \) \(=\) \(\ds \int_0^\infty \frac 0 {x^3} \rd x\) Sine of Zero is Zero
\(\ds \) \(=\) \(\ds 0\)

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:

\(\ds \map {I'} \alpha\) \(=\) \(\ds \frac \d {\d \alpha} \int_0^\infty \frac {\map {\sin^3} {\alpha x} } {x^3} \rd x\)
\(\ds \) \(=\) \(\ds \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {\map {\sin^3} {\alpha x} } {x^3} } \rd x\) Definite Integral of Partial Derivative
\(\ds \) \(=\) \(\ds \frac 1 4 \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {3 \map \sin {\alpha x} - \map \sin {3 \alpha x} } {x^3} } \rd x\) Cube of Sine
\(\ds \) \(=\) \(\ds \frac 3 4 \int_0^\infty \frac {\map \cos {\alpha x} - \map \cos {3 \alpha x} } {x^2} \rd x\) Derivative of Cosine Function, Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac 3 4 \times \frac \pi 2 \paren {3 \alpha - \alpha}\) Definite Integral to Infinity of $\dfrac {\cos p x - \cos q x} {x^2}$
\(\ds \) \(=\) \(\ds \frac {3 \alpha \pi} 4\)

We therefore have:

\(\ds \int_0^1 \map {I'} \alpha \rd \alpha\) \(=\) \(\ds \intlimits {\frac {3 \alpha^2 \pi} 8} 0 1\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac {3 \pi} 8\)

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:

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

$\blacksquare$


Sources

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