Proof of Integration formula $$\int_0^{\infty}x^{-1}e^{-ax}\sin (bx) \;\mathrm dx = \arctan \frac{b}{a}$$
How to prove this result?
 A: Let
$$F(a)=\int_0^{\infty}x^{-1}e^{-ax}\sin (bx) \;\mathrm dx $$
then we can prove using Leibniz theorem: differentiate under the sign $\int$ that:
$$F'(a)=-\int_0^{\infty}e^{-ax}\sin (bx) \;\mathrm dx=-\operatorname{Im} \int_0^{\infty}e^{(-a+ib)x} \;\mathrm dx=\operatorname{Im}\frac{1}{-a+ib}=-\frac b{a^2+b^2}$$
so 
$$F(a)=-\int \frac b{a^2+b^2}da=\arctan\frac b a+C$$
Notice that $C=0$ since the integral is zero for $b=0$.
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}x^{-1}\expo{-ax}\sin\pars{bx}\,\dd x
     = \arctan\pars{b \over a}:\ {\large ?}}$

Assumming $\ds{a > 0}$:
  \begin{align}
&\color{#00f}{\large\int_{0}^{\infty}x^{-1}\expo{-ax}\sin\pars{bx}\,\dd x}=
\sgn\pars{b}\int_{0}^{\infty}\exp\pars{-\,{a \over \verts{b}}\,x}\,
{\sin\pars{x} \over x}\,\dd x
\\[3mm]&=
\sgn\pars{b}\int_{0}^{\infty}\exp\pars{-\,{a \over \verts{b}}\,x}\,
\pars{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}\,\dd x
\\[3mm]&=
\half\sgn\pars{b}\int_{-1}^{1}\braces{\int_{0}^{\infty}
\exp\pars{\bracks{-\,{a \over \verts{b}} + \ic k}x}\,\dd x}\,\dd k
=
\half\sgn\pars{b}\int_{-1}^{1}{1 \over a/\verts{b} - \ic k}\,\dd k
\\[3mm]&=
\sgn\pars{b}\int_{0}^{1}{a/\verts{b} \over \pars{a/b}^{2} + k^{2}}\,\dd k
=
\sgn\pars{b}\sgn\pars{a}\int_{0}^{\verts{b/a}}{1 \over k^{2} + 1}\,\dd k
\\[3mm]&=\sgn\pars{b \over a}\arctan\pars{\verts{b \over a}}
=\color{#00f}{\large\arctan\pars{b \over a}}
\end{align}

A: Use Parseval's Theorem:
$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k)$$
where $f$ and $F$ are Fourier transform pairs, as are $g$ and $G$.  We identify
$$f(x) = e^{-a x} \theta(x) \implies F(k) = \frac1{a-i k}$$
$$g(x) = \frac{\sin{b x}}{x} \implies G(k) = \begin{cases}\pi & |k| \le b \\ 0 & |k| \gt b \end{cases}$$
where $\theta(x)$ is the Heaviside step function ($0$ when $x \lt 0$, $1$ when $x \gt 0$).Then the integral is
$$\frac12 \int_{-b}^b \frac{dk}{a-i k} = \frac{i}{2} \log{\left (\frac{1-i b/a}{1+i b/a} \right )} = \arctan{\frac{b}{a}}$$
