How compute Inverse Fourier of this function $\mathcal{F}^{-1} \Big\{\frac{1}{iw - \alpha^2 k^2}\Big\}$ I need to compute this Inverse Fourier Transform to arrive at the given result
$$
p(t) = \mathcal{F}^{-1} \Big\{\frac{1}{iw - k^2\alpha^2 }\Big\} = -\sqrt{2\pi}H(t)e^{-k^2\alpha^2t}
$$
Where $H(t)$ is the Heaviside Function
I really don't have idea how can i start to tackle this problem. Any hint will be apreciated :)
 A: Observe that from the Residue Theorem, we have for $t>0$
$$\begin{align}
\oint_C \frac{e^{-iz t}}{iz-k^2\alpha^2}\,dz&=\int_{-R}^{R}\frac{e^{-i\omega t}}{i\omega-k^2\alpha^2}\,d\omega+\int_{0}^{-\pi}\frac{e^{-iRe^{i\theta} t}}{iRe^{i\theta} -k^2\alpha^2}\,iRe^{i\theta}d\theta\\\\
=-2\pi i \frac{e^{k^2\alpha^2 t} }{i}\\\\
=-2\pi e^{k^2\alpha^2 t} \tag 1
\end{align}$$
having closed the contour in the lower-half plane, while for $t<0$ we have
$$\begin{align}
\oint_C \frac{e^{-iz t}}{iz-k^2\alpha^2}\,dz&=\int_{-R}^{R}\frac{e^{-i\omega t}}{i\omega-k^2\alpha^2}\,d\omega+\int_{0}^{\pi}\frac{e^{iRe^{i\theta} t}}{iRe^{i\theta} -k^2\alpha^2}\,iRe^{i\theta}d\theta\\\\
=0 \tag 2
\end{align}$$
having closed the contour in the upper half plane.
As $R\to \infty$ the second integrals on the right-hand sides of $(1)$ and $(2)$ vanish, while the first integrals becomes the integral of interest.  Thus
$$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{e^{-i\omega t}}{i\omega-k^2\alpha^2}\,d\omega=
\begin{cases}
-\sqrt{2\pi} e^{k^2\alpha^2 t}&,t>0\\\\
0&,t<0
\end{cases}$$
as was to be shown!
