Inverse Fourier Transform For Solution Of Diffusion Equation With Convection My Question:
How do I complete the inverse Fourier Transform of:
$\displaystyle \int_{-\infty}^\infty F(\omega)e^{-k\omega^2t}e^{-ci\omega t}e^{-i\omega x}\,dx$
I cant figure out quite how to use the shift/convolution theorem here.
The Problem:

Solve the diffusion equation with convection:
$\displaystyle \frac{\partial u}{\partial t} = k\frac{\partial^2
 u}{\partial x^2}+c\frac{\partial u}{\partial x}, -\infty \lt x \lt
 \infty$
$\displaystyle u(x,0)=f(x)$

What I have done so far:
$\displaystyle \mathcal{F}\left[\frac{\partial u}{\partial t}\right] = k\mathcal{F}\left[\frac{\partial^2 u}{\partial x^2}\right]+c\mathcal{F}\left[\frac{\partial u}{\partial x}\right]$
...
$\displaystyle \frac{dU}{dt}=-k\omega^2U-ci\omega U$
$\displaystyle \implies U(\omega,t)=C(\omega)e^{-k\omega^2t}e^{-ci\omega t}$
$\displaystyle u(x,0)=f(x)\implies U(\omega,t)=F(\omega)e^{-k\omega^2t}e^{-ci\omega t}$
Let $\displaystyle G(\omega)=e^{-k\omega^2t}e^{-ci\omega t}$, and $H(\omega)=F(\omega)G(\omega)$
Then $U(\omega,t)=H(\omega)$
$\mathcal{F}^{-1}[U(\omega,t)]=\mathcal{F}^{-1}[H(\omega)]$
$\displaystyle \implies u(x,t)=h(x)$
$\displaystyle = \int_{-\infty}^\infty F(\omega)G(\omega)e^{-i\omega x}\,dx$
$\displaystyle = \int_{-\infty}^\infty F(\omega)e^{-k\omega^2t}e^{-ci\omega t}e^{-i\omega x}\,dx$ (Stuck here, when trying to do the inverse....)

Suppose $u(t, x)$ a solution to the original pde, then define $v(t, x)= e^{-ct}u(t, x)$. Observe
\begin{align}
v_t-k v_{xx} =&\ -ce^{-ct}u(t, x)+e^{ct}u_t(t, x)-ke^{ct}u_{xx}(t, x) \\
=&\ e^{ct}(u_t(t, x)-ku_{xx}(t, x)-cu(t, x)) = 0
\end{align}
which is just the heat equation. Now solve it with Fourier transform or whatever you like. Then we see that
\begin{align}
u(t, x) = e^{ct}v(t, x)
\end{align}
is your solution. 
