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

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

Proof:
From Definite Integral 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:
{{begin-eqn}}
{{eqn | l = \int_{-\infty}^\infty \map f x \rd x
      | r = \int_{-\infty}^\infty \frac {x \cos m x} {x^2 + a^2} \rd x + i \int_{-\infty}^\infty \frac {x \sin m x} {x^2 + a^2} \rd x
      | c = Linear Combination of Definite Integrals
}}
{{eqn | r = i \int_{-\infty}^\infty \frac {x \sin m x} {x^2 + a^2} \rd x
      | c = Definite Integral of Odd Function
}}
{{end-eqn}}
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:
{{begin-eqn}}
  {{eqn  | l = \int_{\Gamma} \map f x \rd x
         | r = \int_{C_1} \frac {x e^{i m x} } {x^2 + a^2} \rd x + \int_{C_2} \frac {x e^{i m x} } {x^2 + a^2} \rd x
         | c = Contour Integral of Concatenation of Contours
}}
{{eqn    | r = \int_{-R}^R \frac {x e^{i m x} } {x^2 + a^2} \rd x + \int_{C_2} \frac {x e^{i m x} } {x^2 + a^2} \rd x
         | c = {{Defof|Complex Contour Integral}}
}}
{{end-eqn}}
The integral over $C_2$ can be shown to vanish as $R \to \infty$:
{{begin-eqn}}
{{eqn | l = \size {\int_{C_2} \frac {x e^{i m x} } {x^2 + a^2} \rd x}
      | o = \le
      | r = \frac \pi m \max_{0 \le \theta \le \pi} \size {\frac {R e^{i \theta} } {R^2 e^{2 i \theta} + 1} }
      | c = Jordan's Lemma
}}
{{eqn | r = \frac \pi m \paren {\frac R {R^2 - 1} } 
}}
{{eqn | o = \to
      | r = 0
      | c = as $R \to \infty$
}}
{{end-eqn}}
Taking $R \to \infty$, we have:
{{begin-eqn}}
{{eqn  | l = \int_{-\infty}^\infty \frac {x e^{i m x} } {x^2 + a^2} \rd x
       | r = \int_{C_1} \frac {x e^{i m x} } {x^2 + a^2} \rd x + 0
       | c = The integral over $C_2$ vanished
}}
{{eqn  | r = 2 \pi i \Res {\frac {x e^{i m x} } {x^2 + a^2} } {a i}
       | c = Cauchy's Residue Theorem
}}
{{eqn  | r = 2 \pi i \paren {\lim_{x \mathop \to ai} \paren {x - a i} \frac {x e^{i m x} } {\paren {x + a i} \paren {x - a i} } }
       | c = Residue at Simple Pole
}}
{{eqn  | r = 2 \pi i \paren {\frac {x e^{i m x} } {x + a i} }_{x = a i}
       | c = 
}}
{{eqn  | r = \pi i e^{-m a}
}}
{{end-eqn}}
So: 
:$\ds \int_{-\infty}^\infty \frac {x \sin m x} {x^2 + a^2} \rd x = \pi e^{-m a}$
giving:
:$\ds \int_0^\infty \frac {x \sin m x} {x^2 + a^2} \rd x = \frac \pi 2 e^{-m a}$
{{qed}}