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
With a view to expressing the primitive in the form:
 * $\ds \int f g' \rd t = f g - \int f' g \rd t$

let:

So:

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

Then:

Adjoin to the case where $p = q$.

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

Then:

Adjoin to the case where $p = q$.