Definite Integral to Infinity of x by Sine m x over x Squared plus a Squared/Proof 2

Theorem

$\ds \int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x = \frac \pi 2 e^{-m a}$

Proof

From Definite Integral to Infinity of $\dfrac {\cos m x} {x^2 + a^2}$:

$\ds \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x = \frac \pi {2 a} e^{-m a}$

We have:

\(\ds \frac \d {\d m} \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x\)

\(=\)

\(\ds \int_0^\infty \frac \partial {\partial m} \paren {\frac {\cos m x} {x^2 + a^2} } \rd x\)

Definite Integral of Partial Derivative

\(\ds \)

\(=\)

\(\ds -\int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x\)

Derivative of $\cos a x$

So:

\(\ds \int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x\)

\(=\)

\(\ds -\frac \d {\d m} \paren {\frac \pi {2 a} e^{-m a} }\)

\(\ds \)

\(=\)

\(\ds \frac \pi 2 e^{-m a}\)

Derivative of $e^{a x}$

$\blacksquare$