Fourier transform of $e^{-iwt}$ First, I have searched the internet as well as here for an answer to my question but did not find one (or anything close enough to lead me to the answer on my own). 
I have a basic understanding of how the Fourier transform works. I know that there are specific transforms and properties that can be combined in order to transform some function. I have looked over some derivations of Fourier transform pairs but not all of them.
I am trying evaluate the following integral $$G\left(\omega\right)=\int_{-\infty}^{\infty}e^{-i\omega t}\left(e^{-i\omega t}g\left(t\right)\right)dt$$ where $g\left(t\right)$ is left general and $\omega$ is non-constant in both exponents.
 A: We actually don't need to do any special systematic integration to figure this one out - intuitive, physical reasoning will suffice.
Remember that the Fourier transform of a function, intuitively, gives you the "density" of energy within the graph of the function, if thought of as a time-series signal, for each frequency component. Equivalently, it is a distribution function for the energy amongst the frequencies in the frequency spectrum.
Now,
$$f(t) := e^{-i\omega_0 t}$$
has only a single frequency component with angular frequency $-\omega_0$. Thus all of its energy is concentrated there. Hence, the Fourier transform should be zero at every other angular frequency except the one named above, i.e. $-\omega_0$, and at that frequency the only reasonable option is to have a delta spike - a "point mass" charged with all the energy in the signal. Thus,
$$F(\omega) = \delta\left(\omega - (-\omega_0)\right)$$
so technically a pseudo-function. The last step, then, is to prove mathematically that this is the correct solution by taking the inverse Fourier transform:
$$\mathcal{F}^{-1}[F](t) = \int_{-\infty}^{\infty} F(\omega)\ e^{+i\omega t}\ d\omega$$
which gives
$$\begin{align}
\mathcal{F}^{-1}[F](t) &= \int_{-\infty}^{\infty} F(\omega)\ e^{+i\omega t}\ d\omega\\
&= \int_{-\infty}^{\infty} \delta\left(\omega - (-\omega_0)\right) e^{+i\omega t}\ d\omega \\
&= e^{+i (-\omega_0) t}\\
&= e^{-i\omega_0 t}\\
&= f(t)\end{align}$$
simply using the defining property of the delta function: whenever you integrate
$$\int_{-\infty}^{\infty} \delta(x - x_0)\ h(x)\ dx$$
for any function $h(x)$, the result of the integral is always $h(x_0)$, in effect, "by definition". Hence setting $h(x) := e^{+i x t}$ (using $x$ instead of $\omega$ and $x_0$ instead of $\omega_0$), you get the result above, and so the given $F$ is, indeed, the Fourier transform of $f$.
A: we know,
$\mathscr{F}\{g(t)\}=G(i\omega)$
by frequency shifting property i.e, multiplication by exponential in time domain leads to translation in frequency domain
so,
$\mathscr{F}\{e^{-iwt}.g(t)\}=\displaystyle\int_{-\infty}^{\infty}(e^{-iwt}g(t))e^{-i\omega t}dt=G(i(\omega+w))$
where $w$ is scalar and is different from $\omega $ (omega) which is frequency in (radian per second )
