Definite Integral to Infinity of x by Sine m x over x Squared plus a Squared/Proof 1

Proof
From Definite Integral of Even Function:


 * $\ds \frac 1 2 \int_{-\infty}^\infty \frac {x \sin m x} {x^2 + a^2} \rd x = \int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x$

With the aim of integrating over the domain, we split the domain up into $2$ components as follows:

Let $R$ be a positive real number with $R > a$.

Let $C_1$ be the straight line segment from $-R$ to $R$.

Let $C_2$ be the arc of the circle of radius $R$ centred at the origin connecting $R$ and $-R$ anticlockwise.

Let $\Gamma = C_1 \cup C_2$.

Let:


 * $\map f x = \dfrac {x e^{i m x} } {x^2 + a^2}$

From Euler's Formula, we have:


 * $\map f x = \dfrac {x \cos m x} {x^2 + a^2} + i \dfrac {x \sin m x} {x^2 + a^2}$

So:

Note that the integrand is meromorphic with simple poles where $x^2 + a^2 = 0$.

That is, at $x = a i$ and $x = -a i$.

As our semi-circular contour lies in the upper half-plane, the only pole of concern is $a i$.

As $R > a$, these poles do not lie on $C_2$, but are enclosed by the curve $\Gamma$.

We have:

The integral over $C_2$ can be shown to vanish as $R \to \infty$:

Taking $R \to \infty$, we have:

So:


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

giving:


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