Suppose that a function$\,$ $f(t)$ $\,$has the Fourier transform $\,$$\hat f(\omega)=e^{-\omega^4}$. What general rules are there for computing the Fourier transforms of things like $\,$$g(t)=f(2t)$, $\,$ $h(t)=f(2t+1)$, $k(t)=f(2t+1)e^{it}$ and similar?
I know that the Fourier transforms are the once below and I know how to find them (by finding f(t) by the inverse transform and then transforming f(2t+1) etc.) My question is if there is a quicker way of finding a Fourier transform of a function with a modified argument if the transform of the function is known.(?)
$$\hat g (\omega)=\frac{1}{2} \hat f \left(\frac{\omega}{2}\right)=\frac{1}{2}e^{-\frac{\omega^4}{16}}$$
$$\hat h (\omega)=e^{i \frac{\omega}{2}} \hat g (\omega) = \frac{1}{2}e^{i \frac{\omega}{2}-\frac{\omega^4}{16}}$$
$$\hat k (\omega)= \hat h (\omega-1)=\frac{1}{2}e^{i \frac{(\omega-1)}{2}-\frac{(\omega-1)^4}{16}}$$
There are indeed such laws!
Throughout, we will consider ourselves with a function $f(t)$ and its Fourier transform, and we will let $\gamma$ be a real constant.
We will also suppose $f$ has a defined transform and inverse transform where needed, denoted $\mathscr{F} \left. [f(t)] \right|_\omega$ and $\mathscr{F}^{-1} \left. [f(\omega)] \right|_t$ respectively.
Scalar Laws: (we further assume $\gamma \neq 0$ here)
$$\mathscr{F} \left. [f(\gamma t)] \right|_\omega = \frac{1}{|\gamma |} \mathscr{F} \left. [f(t)] \right|_\omega$$
$$\mathscr{F}^{-1} \left. [f(\gamma \omega)] \right|_t = \frac{1}{|\gamma |} \mathscr{F}^{-1} \left. [f(\omega)] \right|_t$$
Translation Laws:
$$\mathscr{F} \left. [f(t - \gamma )] \right|_\omega = e^{-i2\pi \gamma \omega} \cdot \mathscr{F} \left. [f(t)] \right|_\omega$$
$$\mathscr{F}^{-1} \left. [f(\omega - \gamma )] \right|_t = e^{i2\pi \gamma \omega} \cdot \mathscr{F}^{-1} \left. [f(\omega)] \right|_t$$
Equivalently, following directly from these, one can also state,
$$\mathscr{F} \left. [e^{i2\pi \gamma \omega} \cdot f(t)] \right|_\omega = \mathscr{F} \left. [f(t)] \right|_{\omega - \gamma}$$
$$\mathscr{F}^{-1} \left. [e^{-i2\pi \gamma \omega} \cdot f(\omega)] \right|_t = \mathscr{F}^{-1} \left. [f(\omega)] \right|_{t - \gamma }$$
You can utilize these as necessary to manipulate your function into a "nicer" place. You might need to utilize multiple identities in sequence
For example, in finding the transform of $f(2t+1)$, you'd first want to note that $f(2t+1)=f(2(t+\frac{1}{2}))$. Then you can apply a translation identity with $\gamma = -\frac{1}{2}$, which gives you an $e^{(stuff)}$ factor times the transform of $f(2t)$. Then you apply the scaling identity to get a constant times the transform of $f(t)$.
Or, to just show the calculation, first we use the translation identity:
$$\mathscr{F} \left. [f(2t+1)] \right|_\omega = \mathscr{F} \left. \left[ f \left( 2 \left( t+\frac{1}{2} \right) \right) \right] \right|_\omega = e^{-i2\pi \left( - \frac{1}{2} \right) \omega} \cdot \mathscr{F} \left. [f(2t)] \right|_\omega = e^{i\pi \omega} \cdot \mathscr{F} \left. [f(2t)] \right|_\omega$$
Then we use the scaling identity on the transform we got:
$$\mathscr{F} \left. [f(2t)] \right|_\omega = \frac{1}{2} \cdot \mathscr{F} \left. [f(t)] \right|_\omega$$
Thus, overall,
$$\mathscr{F} \left. [f(2t+1)] \right|_\omega = \frac{1}{2} \cdot e^{i\pi \omega} \cdot \mathscr{F} \left. [f(t)] \right|_\omega$$