Use contour integration to compute the Fourier transform, The problem statement is:
Use contour integration to determine the Fourier transform,
$\large \hat f(ξ)=∫_{-\infty}^{\infty}f(x)e^{−iξx}dx$,
of $\large f(x)=\frac{1}{2−2x−x^2}$.
Some issues that I am running into are:
Attempt #1
I originally chose a lower semi-circular contour, since the negative i on the exponential function will help give the needed estimate of zero along the semi-circular contour, as we let it grow to infinity - on the upper semi-circular contour, the positive imaginary axis poses problems for the factor $e^{-i\xi x}$.
But then the two simple poles of the function are lying on the real axis.
So I had thought of perhaps, when integrating along the real axis, perhaps jump over the two poles, and estimate the integrand on those two tiny circles and show it goes to zero.
But then, my contour would enclose...no poles...and my function would be analytic on and inside the closed contour, and by Cauchy's theorem, my integration yields zero.
Attempt #2
Now, in hopes of enclosing the two (or at least one?) poles, I chose a box contour instead.  A box of height 8, so 4 units above and below the real axis.  This is a square box to start, and ensures that both poles are enclosed.  I then parameterized fixing the width of the box and hoping to let the length of the box grow to infinity.  Then perhaps, integration along the top and bottom edges of this box, which will be integration with respect to the x variable, will give me two desired integrals.  Unfortunately, it looks like these desirable integrals are canceling each other out (of course, by the parametrization of the box).
Now I'm not sure what to do.  
Any hints or suggestions are welcome.
Thanks,
 A: Consider for $\xi \lt 0$ the contour integral
$$\oint_C dz \frac{e^{-i \xi z}}{2-2 z-z^2} $$
where $C$ is a semicircle of radius $R$ in the upper half plane, with a small semicircular indent into the upper half-plane of radius $\epsilon$ at each pole $z_{\pm}=-1 \pm \sqrt{3}$.  The contour integral is then
$$PV \int_{-R}^R dx \frac{e^{-i \xi x}}{2-2 x-x^2} +i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{-i \xi (z_-+\epsilon e^{i \phi})}}{3-(-\sqrt{3}+\epsilon e^{i \phi})^2}\\ + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{-i \xi (z_++\epsilon e^{i \phi})}}{3-(\sqrt{3}+\epsilon e^{i \phi})^2}+ i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{e^{-i \xi R e^{i \theta}}}{2-2 R e^{i \theta} - R^2 e^{i 2 \theta}}$$
In the limit as $R \to \infty$, the fourth integral vanishes as $1/R^2$.  As $\epsilon \to 0$, the sum of the second and third integrals converge to
$$i \frac{\pi}{2 \sqrt{3}} \left (-e^{-i \xi z_-} + e^{-i \xi z_+} \right ) $$
By Cauchy's theorem, the contour integral is equal to zero.  Thus, for $\xi \lt 0$ we have
$$PV \int_{-\infty}^{\infty} dx \frac{e^{-i \xi x}}{2-2 x-x^2} = -\frac{\pi}{\sqrt{3}} e^{i \xi} \sin{\left (\sqrt{3} \xi \right )} $$
For $\xi \gt 0$, one uses a contour in the lower half plane.  
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\hat{\on{f}}\pars{\xi} & \equiv
\bbox[5px,#ffd]{%
\mrm{P.V.}\int_{-\infty}^{\infty}{\expo{−\ic\xi x} \over
2 - 2x - x^{2}}\,\dd x} =
-\expo{\ic\xi}\,\mrm{P.V.}
\int_{-\infty}^{\infty}{\expo{−\ic\xi x} \over x^{2} - 3}\,\dd x
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\mrm{P.V.}
\int_{-\infty}^{\infty}\expo{−\ic\xi x}\pars{%
{1 \over x - \root{3}} - {1 \over x + \root{3}}}\,\dd x
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\sum_{\sigma = \pm}
\sigma\,\mrm{P.V.}\int_{-\infty}^{\infty}
{\expo{−\ic\xi x} \over x - \sigma\root{3}}\,\dd x
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\sum_{\sigma = \pm}
\sigma\left[%
\int_{-\infty}^{\infty}
{\expo{−\ic\xi x} \over x - \sigma\root{3} + \ic 0^{+}}\,\dd x
\right.
\\[2mm] &\ \phantom{-{\root{3} \over 6}\expo{\ic\xi}\,\sum_{\sigma = \pm}
\sigma\left[\right.}
+ \left.\int_{-\infty}^{\infty}
\expo{−\ic\xi x}\,\ic\pi\,\delta\pars{x - \sigma\root{3}}\,\dd x
\right]
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\sum_{\sigma = \pm}
\sigma\bracks{%
-\bracks{\xi < 0}2\pi\ic\expo{−\ic\xi\pars{\sigma\root{3}}} +
\ic\pi\expo{−\ic\xi\sigma\root{3}}}
\\[5mm] = &\
-{\root{3} \over 6}\expo{\ic\xi}\,\pi\ic\
\overbrace{\braces{-2\bracks{\xi < 0} + 1}}
^{\ds{\on{sgn}\pars{\xi}}}\ 
\overbrace{%
\sum_{\sigma = \pm}\sigma\expo{−\ic\xi\pars{\sigma\root{3}}}}
^{\ds{-2\ic\sin\pars{\root{3}\xi}}}
\\[5mm] = &\
\bbx{-{\root{3} \over 3}
\,\pi\expo{\ic\xi}\sin\pars{\root{3}\verts{\xi}}} \\ &
\end{align}
