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

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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:

\(\ds \map {\frac \d {\d x} } {\sin^2 p x}\) \(=\) \(\ds \map {\frac \d {\d x} } {\sin p x} \map {\frac \d {\map \d {\sin p x} } } {\sin^2 p x}\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds 2 p \cos p x \sin p x\) Derivative of Sine of a x, Derivative of Power
\(\ds \) \(=\) \(\ds p \sin 2 p x\) Double Angle Formula for Sine

We also have, by Primitive of Power:

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

So:

\(\ds \int_0^\infty \paren {\frac {\sin p x} x}^2 \rd x\) \(=\) \(\ds \intlimits {-\frac {\sin^2 p x} x} 0 \infty - \int_0^\infty \paren {-\frac {p \sin 2 p x} x} \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds - \lim_{x \mathop \to \infty} \paren {\frac {\sin^2 p x} x} + \lim_{x \mathop \to 0} \paren {\frac {\sin^2 p x} x} + p \int_0^\infty \frac {\sin 2 p x} x \rd x\)


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:

\(\ds \lim_{x \mathop \to 0} \frac {\sin p x} x\) \(=\) \(\ds \lim_{p x \mathop \to 0} \frac {\sin p x} {\frac {p x} p}\)
\(\ds \) \(=\) \(\ds p \lim_{u \mathop \to 0} \frac {\sin u} u\) letting $u = p x$
\(\ds \) \(=\) \(\ds p\) Limit of $\dfrac {\sin x} x$ at Zero

So:

\(\ds \lim_{x \mathop \to 0} \frac {\sin^2 p x} x\) \(=\) \(\ds \paren {\lim_{x \mathop \to 0} \sin p x} \paren {\lim_{x \mathop \to 0} \frac {\sin p x} x}\) Product Rule for Limits of Real Functions
\(\ds \) \(=\) \(\ds p \map \sin {0 p}\)
\(\ds \) \(=\) \(\ds 0\) Sine of Zero is Zero


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.

$\blacksquare$


Sources

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