$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{\infty}{r\sin\pars{r\rho} \over r^{2} - \alpha^{2}}\,\dd r:\
{\large ?}.\quad}$
These integrals are very common in Quantum Mechanics as related to scattering theory. In that cases, $\ds{\alpha}$ has a 'small' imaginary part which indicates whether a wave is going to a scattering center or is leaving the scattering center.
The 'history' is more or less like this:
\begin{align}
&\color{#c00000}{\int_{0}^{\infty}{r\sin\pars{r\rho}\over
r^{2} - \pars{\verts{\alpha} \pm \ic 0^{+}}^{2}}\,\dd r}
\\[3mm]&=\int_{0}^{\infty}r\sin\pars{r\rho}\,
{1 \over 2\pars{\verts{\alpha} \pm \ic 0^{+}}}\bracks{%
{1 \over r - \pars{\verts{\alpha} \pm \ic 0^{+}}}
-{1 \over r + \pars{\verts{\alpha} \pm \ic 0^{+}}}}\,\dd r
\\[3mm]&={1 \over 2\pars{\verts{\alpha} \pm \ic 0^{+}}}\,2\ic\,
\Im\int_{0}^{\infty}{r\sin\pars{r\rho} \over r - \verts{\alpha} \mp \ic 0^{+}}\,\dd r
\\[3mm]&={\ic \over \verts{\alpha} \pm \ic 0^{+}}
\int_{0}^{\infty}r\sin\pars{r\rho}
\bracks{\pm\,\pi\,\delta\pars{r - \verts{\alpha}}}\,\dd r
={\ic \over \verts{\alpha} \pm \ic 0^{+}}
\,\bracks{\pm\pi\verts{\alpha}\sin\pars{\verts{\alpha}\rho}}
\\[3mm]&=\pm\ic\pi\verts{\alpha}\sin\pars{\verts{\alpha}\rho}\,
\bracks{\pp{1 \over \verts{\alpha}} \mp \ic\pi\delta\pars{\verts{\alpha}}}
\end{align}
where $\ds{\delta\pars{x}}$ is the
Dirac Delta 'Function' and we used the identities ( 'under' the integral sign )
$\ds{{1 \over x \pm \ic 0^{+}} = \pp{1 \over x} \mp \ic\pi\delta\pars{x}}$. $\ds{\pp}$ denotes the Principal Value.
$$
\color{#00f}{\large\int_{0}^{\infty}
{r\sin\pars{r\rho} \over r^{2} - \pars{\verts{\alpha} \pm \ic 0^{+}}^{2}}\,\dd r
=\pm\,\ic\pi\sin\pars{\verts{\alpha}\rho}}
$$