This is not a complete solution, but may be one reasonable approach to it.
Firts define:
$$I(a)=\int_{0}^{\infty} \frac{\cos(ax)}{{\left(1+x^{2}\right)}^{c}}dx$$
Which can itself be written in terms of $K_{v}(x)$
Taking the derivative of $I(a)$ with respect to a gives us
$$\frac{dI(a)}{da}=-\int_{0}^{\infty} \frac{x\sin(ax)}{{\left(1+x^{2}\right)}^{c}}dx$$
Now, consider the following integral, easily verified by a change of variable
$$\Gamma(c)={\left(1+x^{2}\right)}^{c}\int_{0}^{\infty}e^{-\left(1+x^{2}\right)u} u^{c-1}du$$
now multiply $I(a)$ by $\Gamma(c)$
$$\Gamma(c)I(a)=\int_{0}^{\infty}cos(ax)\int_{0}^{\infty}e^{-\left(1+x^{2}\right)u} u^{c-1}dudx$$
Swap the integrals and distribute the exponential
$$\Gamma(c)I(a)=\int_{0}^{\infty}u^{c-1}e^{-u}\int_{0}^{\infty}e^{-x^{2}u}\cos(ax)dxdu$$
The inner integral has the following solution
$$\frac{1}{2}\sqrt{\frac{\pi}{u}}e^{-\frac{a^{2}}{4u}}$$
Giving us:
$$\Gamma(c)I(a)=\frac{\sqrt{\pi}}{2}\int_{0}^{\infty}u^{c-\frac{2}{2}}e^{-u-\frac{a^{2}}{4u}}du$$
Make the following substitution $u=\left(\frac{x}{2}\right)e^{v}$
$$
I(a)=\frac{\sqrt{\pi}}{2}\frac{1}{\Gamma(c)}{\left(\frac{a}{2}\right)}^{c-\frac{1}{2}}\int_{-\infty}^{\infty}e^{-a\cosh(v)}e^{\left(c-\frac{1}{2}\right)v} dv$$
because $e^{\left(c-\frac{1}{2}\right)v} = \cosh\left(\left(c-\frac{1}{2}\right)v\right)+\sinh\left(\left(c-\frac{1}{2}\right)v\right)$, and since $\sinh\left(\left(c-\frac{1}{2}\right)v\right)$ is an odd function of $v$
gives us the final result:
$$\boxed{I(a)=\frac{\sqrt{\pi}}{\Gamma(c)}{\left(\frac{a}{2}\right)}^{c-\frac{1}{2}}\int_{0}^{\infty}e^{-a\cosh(v)}\cosh{\left(\left(c-\frac{1}{2}\right)v\right)} dv}$$
$$\boxed{I(a)=\frac{\sqrt{\pi}}{\Gamma(c)}{\left(\frac{a}{2}\right)}^{c-\frac{1}{2}}K_{c-\frac{1}{2}}(x)}$$
The last integral is an Integral representation 2 of $K_{v}(x)$
Now, take the derivative of this expression with respect to $a$.
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[10px,#ffd]{\left.\int_{0}^{\infty}\bracks{{1 \over \pars{1 + \ic x}^{b}} - {1 \over \pars{1 - \ic x}^{b}}}\sin\pars{ax}\,\dd x \,\right\vert_{\ a, b\ >\ 0} = -\,{a^{b - 1}\expo{-a} \over \Gamma\pars{b}}\,\pi\ic}:\ {\Large ?}}$.
\begin{align} &\bbox[10px,#ffd]{\left.\int_{0}^{\infty} \bracks{{1 \over \pars{1 + \ic x}^{b}} - {1 \over \pars{1 - \ic x}^{b}}}\sin\pars{ax}\,\dd x\,\right\vert_{\ a,b\ >\ 0}} \\[5mm] = &\ \ic\,\Im\int_{-\infty}^{\infty}{\sin\pars{ax} \over \pars{1 + \ic x}^{b}}\,\dd x \\[2mm] = &\ \ic\,\Im\int_{-\infty}^{\infty}\sin\pars{ax}\ \overbrace{\bracks{{1 \over \Gamma\pars{b}}\int_{0}^{\infty}t^{b - 1} \expo{-\pars{1 + \ic x}t}\dd t}} ^{\ds{=\ {1 \over \pars{1 + \ic x}^{b}}}}\,\dd x \\[5mm] = &\ {\ic \over \Gamma\pars{b}}\,\Im\int_{0}^{\infty}t^{b - 1}\expo{-t} \int_{-\infty}^{\infty}\bracks{\expo{-\pars{t - a}\ic x} - \expo{-\pars{t + a}\ic x}\over 2\ic}\dd x\,\dd t \\[5mm] = &\ -\,{\ic\pi \over \Gamma\pars{b}}\int_{0}^{\infty}t^{b - 1}\expo{-t} \bracks{\delta\pars{t - a} - \delta\pars{t + a}}\,\dd t \\[5mm] = &\ \bbx{-\,{a^{b - 1}\expo{-a} \over \Gamma\pars{b}}\,\pi\ic} \qquad\qquad a, b > 0 \\[5mm] &\ \mbox{} \end{align}