Integral to Infinity of Square of Sine p x over x Squared

Theorem

 * $\ds \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x = \frac {\pi \size p} 2$

where $p$ is a real number.

Proof
We have:

We also have, by Primitive of Power:


 * $\ds \int \frac {\d x} {x^2} = -\frac 1 x + C$

So:

We have, from Real Sine Function is Bounded, that:


 * $0 \le \sin^2 p x \le 1$

for all real $x$.

Therefore:


 * $0 \le \dfrac {\sin^2 p x} x \le \dfrac 1 x$

for all strictly positive real $x$.

We have:


 * $\ds \lim_{x \mathop \to \infty} \frac 1 x = 0$

Therefore, by the Squeeze Theorem, we have:


 * $\ds \lim_{x \mathop \to \infty} \paren {\frac {\sin^2 p x} x} = 0$

As to the other limit, note that:

So:

So:


 * $\ds \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x = p \int_0^\infty \frac {\sin 2 p x} x \rd x$

By Integral to Infinity of Sine p x over x, we have:


 * $\ds \int_0^\infty \frac {\sin 2 p x} x \rd x = \begin{cases} \dfrac \pi 2 & : p > 0 \\

0 & : p = 0 \\ -\dfrac \pi 2 & : p < 0 \end{cases}$

So:


 * $\ds p \int_0^\infty \frac {\sin 2 p x} x \rd x = \begin{cases} \dfrac \pi 2 p & : p > 0 \\

0 & : p = 0 \\ -\dfrac \pi 2 p & : p < 0 \end{cases}$

Hence:


 * $\ds p \int_0^\infty \frac {\sin 2 p x} x \rd x = \frac {\pi \size p} 2$

Hence the result.