Calculate the fourier transform of $(ax^2+bx+c)^{-1}$ 
Find the fourier transform of the function $(ax^2+bx+c)^{-1}$ with $a>0$ and $b^2-4ac<0$.

So my idea for this was using the fact that the fourier transform of the function $f(x)=e^{-\lambda|x|}$ ,  is $\hat f(\xi)=\frac{2\lambda}{\lambda^2+4π\xi^2}$ and then using this $$f(x) = \int\limits_{-\infty}^\infty\hat f(\xi)e^{2\pi\mathrm ix\xi}\,\mathrm d\xi.$$
 A: Completing the square,
$$ ax^2 + bx + c = a\bigg( \Big( x + \frac{b}{2a} \Big)^2 + \frac{4ac-b^2}{4a^2} \bigg) $$
For simplicity, suppose that the roots of the quadratic are not real, so we don't have to worry about singularities. Hence the second term in the bracket is positive, so it will suffice to find the Fourier transform of $ f(y) = 1/(y^2 + m^2) $ and apply various shift and scaling properties.
But we know that this is
$$ \mathcal{F}\left( \frac{1}{y^2+m^2} \right)(k) = \mathcal{F}^{-1}\left( \frac{1}{y^2+m^2} \right)(k) = \frac{\pi}{m} e^{-2\pi m\lvert k \rvert} $$
(the function is even, and the Fourier transform with the normalisation you specify is its own inverse for even functions).
Now, we want the transform of $ f(x+b/2)/a$ with $m= \sqrt{4ac-b^2}/(2a)$. If $g(x) = f(x-\alpha)$, $\mathcal{F}(g)(k) = e^{-2\pi i \alpha}\mathcal{F}(f)(k)$, so
$$ \mathcal{F}\left( \frac{1/a}{(x+b/2)^2+m^2} \right)(k) = \frac{2\pi}{\sqrt{4ac-b^2}} \exp\left( \frac{\pi}{a} \big( i - \sqrt{4ac-b^2}\lvert k \rvert \big) \right) $$
A: You may write
$$
    ax^2+bx+c = a\left(x-i\frac{\sqrt{4ac-b^2}}{2a}\right)\left(x+i\frac{\sqrt{4ac-b^2}}{2a}\right).
$$
Let $r=\sqrt{4ac-b^2}/2a$, which is real and positive by assumption. Then
$$ax^2+bx+c=a(x-ir)(x+ir)$$
and
$$
      \frac{1}{ax^2+bx+c} = \frac{1}{2iar}\left(\frac{1}{x-ir}-\frac{1}{x+ir}\right)
$$
Then
$$
      \hat{f}(\xi)=\frac{\pi}{ar}\frac{1}{2\pi i}\int_{-\infty}^{\infty}e^{2\pi i z\xi}\left(\frac{1}{z-ir}-\frac{1}{z+ir}\right)dz
$$
If $\xi > 0$, then $e^{2\pi ix\xi}$ decays as $\Im z\rightarrow \infty$, and the integral may be traded for the limit of a positively-oriented, semicircular contour integral that closes in the upper half-plane, which may be evaluated using the residue at $ir$ to be
$$
         \frac{\pi}{ar}e^{-2\pi r\xi}.
$$
If $\xi < 0$, then $e^{2\pi iz\xi}$ decays as $\Im z \rightarrow -\infty$, and the integral may be traded for the limit of a negatively-oriented, semicircular contour integral that closes in the lower half-plane, which may be evaluated using the residue at $-ir$ to be
$$
                 \frac{\pi}{ar}e^{2\pi r\xi}
$$
Putting the two pieces together for positive and negative real $\xi$ gives
$$
        \hat{f}(\xi) = \frac{\pi}{ar}e^{-2\pi r|\xi|},\;\;\; r=\sqrt{4ac-b^2}/2a
$$
