Integral to Infinity of One minus Cosine p x over x Squared

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: $15.37$