Definite Integral to Infinity of Cosine p x minus Cosine q x over x Squared

Theorem

$\ds \int_0^\infty \frac {\cos p x - \cos q x} {x^2} \rd x = \frac {\pi \paren {\size q - \size p} } 2$

where $p, q$ are real numbers.

Proof

\(\ds \int_0^\infty \frac {\cos p x - \cos q x} {x^2} \rd x\)

\(=\)

\(\ds \int_0^\infty \frac {1 - \cos q x - \paren {1 - \cos p x} } {x^2} \rd x\)

\(\ds \)

\(=\)

\(\ds \int_0^\infty \frac {1 - \cos q x} {x^2} \rd x - \int_0^\infty \frac {1 - \cos p x} {x^2} \rd x\)

\(\ds \)

\(=\)

\(\ds \frac \pi 2 \size q - \frac \pi 2 \size p\)

Integral to Infinity of $\dfrac {1 - \cos p x} {x^2}$

\(\ds \)

\(=\)

\(\ds \frac {\pi \paren {\size q - \size p} } 2\)

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 15$: Definite Integrals involving Trigonometric Functions: $15.39$