Definite Integral to Infinity of Sine m x over x by x Squared plus a Squared

Theorem

 * $\displaystyle \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x = \frac \pi {2 a^2} \paren {1 - e^{-m a} }$

where:
 * $a$ is a positive real number
 * $m$ is a non-negative real number.

Proof
Fix $a$ and set:


 * $\displaystyle \map I m = \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x$

for $m \ge 0$.

We have:

So by Primitive of $e^{a x}$:


 * $\displaystyle \map I m = -\frac \pi {2 a^2} e^{-m a} + C$

for some constant $C \in \R$.

We have:

On the other hand:

So:


 * $C = \dfrac \pi {2 a^2}$

giving:


 * $\displaystyle \map I m = \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x = \frac \pi {2 a^2} \paren {1 - e^{-m a} }$