TITLE: show that $\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}(1-\frac{e^{-ab}(ab+2)}{2})$
QUESTION [7 upvotes]: show that 
$$\int_{0}^{\infty } \frac{\sin (ax)}{x(x^2+b^2)^2}dx=\frac{\pi}{2b^4}\left(1-\frac{e^{-ab}(ab+2)}{2}\right)$$
for $a,b> 0$
I would like someone solve it using contour but also I would to see different solution using different way to solve it
is there any help 
thanks for all

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty }{\sin\pars{ax} \over x\pars{x^{2} + b^{2}}^{2}}\,\dd x} =
{1 \over 2}\on{sgn}\pars{a}\,
\Im\int_{-\infty}^{\infty }{\expo{\ic\verts{a}x} - 1 \over
x\pars{x^{2} + b^{2}}^{2}}\,\dd x
\\[5mm] = &\
{1 \over 2}\on{sgn}\pars{a}\,\Im\braces{%
2\pi\ic\,{1 \over \pars{2 - 1}!}\,
\lim_{x\ \to\ \ic\verts{b}}\,\,\,
\partiald{}{x}\bracks{\expo{\ic\verts{a}x} - 1 \over
x\pars{x + \ic \verts{b}}^{2}}}
\\[5mm] = &\
\bbx{\on{sgn}\pars{a}\,{\pi \over 2b^{4}}
\bracks{1 - {\expo{-\verts{ab}}\pars{\verts{ab} + 2} \over 2}}} \\ &
\end{align}