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

Theorem

 * $\ds \int_0^\infty \frac {\sin p x \sin q x} {x^2} \rd x = \begin {cases} \dfrac {\pi p} 2 & : 0 < p \le q \\

\dfrac {\pi q} 2 & : p \ge q > 0 \end {cases}$

Proof
From Integration by Parts:


 * $\ds \int f g' \rd t = f g - \int f' g \rd t$

Here:

So:

Case $0 < p \le q$:
Suppose $0 < p < q$

Adjoin to case where $p = q$

Case $p \ge q > 0$:
Suppose $p > q > 0$

Adjoin to case where $p = q$