Prove that the integral $\hat f(\xi)= \int _{-\infty} ^{\infty} \ \frac{e^{-ix\xi}}{x^2-2x+10}$ is valid for all $\xi$. We need to evaluate 

$$\hat{f}(\xi)=\int_{-\infty}^{\infty}\dfrac{e^{-ix\xi}}{x^2-2x+10}\mathrm{dx}$$

and we want to show that for all $\xi\in\Bbb R$ this integral is valid. 
I have no problem proving the case when $\xi <0 $ but the other case I can't seem to get my head around. Specifically, I don't understand how I can show that the contribution over the semi circular arc is $0$ in this other case.
Note, I want to use complex integration for this (Residue Theorem, Cauchys Integral formula etc.)
 A: For any $\xi\in\mathbb{R}$,
$$ \left|\int_{-\infty}^{+\infty}\frac{e^{-ix\xi}}{(x-1)^2+9}\right|\leq\int_{-\infty}^{+\infty}\frac{dz}{z^2+9}=\frac{\pi}{3} $$
by the triangle inequality, hence the integral is well-defined. Additionally, assuming $\xi>0$ we have:
$$\int_{-\infty}^{+\infty}\frac{e^{-ix\xi}}{(x-1)^2+9}\,dx=2\pi i\operatorname*{Res}_{x=1+3i}\frac{e^{-ix\xi}}{(x-1)^2+9}=\frac{\pi e^{-\xi i}}{3e^{3\xi}}$$
and in general:
$$\forall \xi\in\mathbb{R},\qquad \int_{-\infty}^{+\infty}\frac{e^{-ix\xi}}{(x-1)^2+9}\,dx= \color{red}{\frac{\pi e^{-\xi i}}{3e^{3|\xi|}}}.$$
A: Okay, so I realised that when we use the parametrization 
$\gamma_{1}= Re^{it}$ with $t \in\ [\pi,2\pi]$ we get that the integral 
$\int_{-\infty}^{+\infty}\frac{e^{-ix\xi}}{(x-1)^2+9}\,dx $ $\to 0$ as we let $R \to \infty$ 
and so all that remains is to calculate 
$ \int_{\gamma_2} \frac{e^{-iz\xi}}{z^2-2z+10}\,dz $
where $\gamma_2 = t \ \ \ t \in [R,-R]$ (this range to ensure proper orientation of the boundary)  and we will get that that is equal to almost what the case $\xi <0$ yields but with 1 little sign change.
$\int_{-\infty}^{+\infty}\frac{e^{-ix\xi}}{x^2-2x+10}\, dx= \dfrac{\pi}{3}e^{-i\xi\pm3\xi}$ { + when $\xi<0$, - when $\xi>0$}
