Proving $\int_{0}^\infty \left(\frac{1}{(1+ix)^b}-\frac{1}{(1-ix)^b}\right)\sin(ax)\mathrm{d}x =\frac{-ia^{b-1}e^{-a}\pi}{\Gamma[b]} $ In Mathematica, $$\int_{0}^\infty \left(\frac{1}{(1+ix)^b}-\frac{1}{(1-ix)^b}\right)\sin(ax)\mathrm{d}x =\frac{-ia^{b-1}e^{-a}\pi}{\Gamma[b]}$$
I want to prove this, but I can't.
If anyone knows the proof of the above definite integral,
Thank you for your instruction.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
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$\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}
A: Consider first
$$I=\int \frac {e^{i a x}}{(1+i x)^b} dx$$ Let $1+ix=y$ to make
$$I=-i \,e^{-a}\int e^{a y}\, y^{-b}\,dy$$  Now, let $ay=-t$
$$\int e^{a y}\, y^{-b}\,dy=(-1)^{b+1} a^{b-1} \int e^{-t} t^{-b}\,dt$$
$$\int e^{-t} t^{-b}\,dt=-\Gamma (1-b,t)$$ I am sure that you can take from here.
