Integral to Infinity of One minus Cosine p x over x Squared
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Theorem
- $\ds \int_0^\infty \frac {1 - \cos p x} {x^2} \rd x = \frac {\pi \size p} 2$
where $p$ is a real number.
Proof
\(\ds \int_0^\infty \frac {1 - \cos p x} {x^2} \rd x\) | \(=\) | \(\ds \int_0^\infty \frac {2 \sin^2 \paren {\frac {p x} 2} } {x^2} \rd x\) | Square of Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \times \frac {\pi \size p} {2 \times 2}\) | Integral to Infinity of $\paren {\dfrac {\sin p x} x}^2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\pi \size p} 2\) |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.37$