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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.36$