I'm interested in the Fourier Spectrum of the solution of the differential equation on the domain $t \in [0, \infty]$:
$$ \frac{dp}{dt} = -\beta p $$
Doing Fourier Analysis on this differential equation I get:
$$ (i \omega + \beta ) P = 0 $$
which is interesting because this would seem to indicate that there are only trivial solutions to this equation.
The time domain solution to this equation is:
$$ p(t) = p(0) e^{-\beta t} \quad \text{for } t \geq 0$$
If I take the fourier transform of this I get:
$$ P (\omega) = \int_{0}^{\infty} p(0) e^{-\beta t} e^{-i\omega t} dt = p(0) \int_{0}^{\infty} e^{-(i\omega + \beta) t} dt $$
$$ P(\omega) = - \frac{p(0)}{i\omega + \beta} e^{-(i\omega + \beta) t} \big |^{t=\infty}_{t = 0} = \frac{p(0)}{i\omega + \beta} $$
Now I plut this into my original equation:
$$ (i \omega + \beta) P = (i \omega + \beta) \frac{p(0)}{i \omega + \beta} = p(0) \neq 0 $$
Where did I go wrong?
In the first case you take the Fourier transform over all of $\mathbb R,$ but in the second case you take it only over the positive half of $\mathbb R.$
If we limit $p(t)$ to be non-zero only for $t>0$ then the first Fourier transform becomes $$ \mathcal{F}\{LHS\} = \int_0^\infty p'(t) \, e^{-i\omega t} \, dt \\ = [p(t) \, e^{-i\omega t}]_0^\infty - \int_0^\infty p(t) \, (-i\omega) e^{-i\omega t} \, dt \\ = -p(0) + i\omega \int_0^\infty p(t) e^{-i\omega t} \, dt \\ = -p(0) + i\omega P(\omega) $$ and $$ \mathcal{F}\{RHS\} = \int_0^\infty (-\beta p(t)) \, e^{-i\omega t} \, dt \\ = -\beta \int_0^\infty p(t) \, e^{-i\omega t} \, dt \\ = -\beta P(\omega) $$
Thus, $$(i\omega+\beta)P(\omega) = p(0)$$ so $$P(\omega) = p(0)/(i\omega+\beta).$$
This is the same as you got in the second case.