How would one take the Fourier Transform of $$ m\frac{d^2x}{dt^2}+m\omega_0^2 x + m\Gamma \frac{dx}{dt} = -eE $$ to get $$ -m\omega^2 x+m\omega_0^2 x + jm\omega\Gamma x = -eE $$
This is in our class lecture notes for nanophotonics, specifically relating to the Lorentz model in plasmonics, but I am having trouble understanding how our professor got from the first expression to the second one.
In addition to the solution given by hyportnex in the comments, you can also see how this relates to the Fourier transform by recalling one of its important properties.
Define the Fourier transform of $x(t)$ as $$X(\omega) \equiv \mathfrak F[x](\omega) \equiv \int_{-\infty}^\infty dt \ e^{-j \omega t} x(t).$$ Now if you do integration by parts, you get $$\mathfrak F[x](\omega) = \underbrace{\frac{j}{\omega} e^{-j\omega t} x(t) \Bigg\vert_{-\infty}^\infty}_0 - \frac{j}{\omega}\int_{-\infty}^\infty dt \ e^{-j\omega t} \ \dot{x}(t).$$ Note that the boundary term is zero because $x$ needs to be a square-integrable function to have a Fourier transform (I'm neglecting the concept of distributions here). You hence get the following important property of the Fourier transform: $$\mathfrak{F}[\dot{x}](\omega) = j \omega \ \mathfrak {F}[x](\omega),$$ which can be generalized for an arbitrary derivative by induction: $$\mathfrak{F}[{x}^{(n)}](\omega) = ( j \omega)^n \ \mathfrak {F}[x](\omega),$$
You can now use this property to derive your second equation. Take the Fourier transform of both sides of $m \ddot{x} + m \Gamma \dot{x} + m \omega_0^2 x = - e E$ to get $$m (j \omega)^2 X(\omega) + m \Gamma (j \omega) X(\omega) + m \omega_0^2 X(\omega) = - e \mathcal{E}(\omega),$$ where $\mathcal E(\omega) \equiv \mathfrak F[E](\omega)$ is the Fourier transform of $E(t)$. This simplifies to $$-m \omega^2 X(\omega) + j \omega m \Gamma X(\omega) + m \omega_0^2 X(\omega) = - e \mathcal{E}(\omega),$$ as desired.
