Evaluate $\int^\infty_{-\infty} \frac{e^{a+ix}}{(a+ix)^b}dx$ where $a>1$ and $b>0.$
I've tried looking at $f(z)=\frac{e^z}{z}$ and integrating around a rectangular contour surrounding the singularity at $z=0$, where the parametrization of one of the side lengths was exactly the integral in question, but I had a hard time calculating the residue at 0 because $b$ is not necessarily an integer.
I also think that maybe the contour should've been a keyhole, since we'll need a branch of the logarithm to even consider $z^b = e^{b\log z}$, but then I confused myself even more. Any hints would be much appreciated.
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\begin{align} &\left. \int_{-\infty}^{\infty}{\expo{a + \ic x} \over \pars{a + \ic x}^{b}}\,\dd x\, \right\vert_{\ a\ >\ 1\,,\ b\ >\ 0} \,\,\,\stackrel{a\ +\ \ic x\ \mapsto\ x}{=}\,\,\, -\ic\int_{a - \infty\ic}^{a + \infty\ic}x^{-b}\expo{x}\,\dd x \\[5mm] &\ \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, \ic\int_{-\infty}^{-\epsilon}\pars{-x}^{-b}\expo{-\pi b\ic}\expo{x}\,\dd x + \ic\int_{\pi}^{-\pi}\epsilon^{-b}\expo{-\ic b\theta}\epsilon\expo{\ic\theta}\ic \,\dd\theta + \ic\int_{-\epsilon}^{-\infty}\pars{-x}^{-b}\expo{\pi b\ic}\expo{x}\,\dd x \\[5mm] = &\ \ic\expo{-\pi b\ic}\int_{\epsilon}^{\infty}x^{-b}\expo{-x}\,\dd x - {2\epsilon^{1 - b}\sin\pars{b\pi} \over b - 1} - \ic\expo{\pi b\ic}\int_{\epsilon}^{\infty}x^{-b}\expo{-x}\,\dd x \\[5mm] = &\ 2\sin\pars{b \pi}\int_{\epsilon}^{\infty}x^{-b}\expo{-x}\,\dd x - {2\epsilon^{1 - b}\sin\pars{b\pi} \over b - 1} \\[5mm] = &\ 2\sin\pars{b \pi}\bracks{% {\epsilon^{1 - b}\expo{-\epsilon} \over b - 1} - \int_{\epsilon}^{\infty}{x^{-b + 1} \over -b + 1}\pars{-\expo{-x}}\,\dd x} - {2\epsilon^{1 - b}\sin\pars{b\pi} \over b - 1} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim} & \,\,\, -\,{2\sin\pars{b \pi} \over b - 1} \int_{\epsilon}^{\infty}{x^{-b + 1}}\expo{-x}\,\dd x \end{align}
In such a case, \begin{align} &\left. \int_{-\infty}^{\infty}{\expo{a + \ic x} \over \pars{a + \ic x}^{b}}\,\dd x\, \right\vert_{\ a\ >\ 1\,,\ b\ >\ 0} = -\,{2\sin\pars{b \pi} \over b - 1}\,\Gamma\pars{-b + 2} \\[5mm] = &\ -\,{2\sin\pars{b \pi} \over b - 1}\, {\pi \over \Gamma\pars{b - 1}\sin\pars{\pi\bracks{b - 1}}} = \bbx{\ds{2\pi \over \Gamma\pars{b}}} \end{align}