Fourier transforms
Hilbert space structure
\[
\int \dd{\omega} e^{i\omega(t_1 - t_2)} = 2 \pi \delta(t_1 - t_2)
\]
Derivation
We verify by integrating against test function $f(t_1,t_2)$ for $\omega \in [-\Omega, \Omega]$\begin{align*}
\int \dd{\Delta t}\dd{\omega} f(t_2 + \Delta t,t_2) e^{i \omega \Delta t}
&= \int \dd{\Delta t} f(t_2 + \Delta t, t_2)
\eval{\frac{e^{i \omega \Delta t}}{i \Delta t}}_{\omega = -\Omega}^{\Omega} \\
&= \int \dd{\Delta t} f(t_2 + \Delta t, t_2) \frac{e^{i \Omega \Delta t}- e^{-i\Omega \Delta t}}{i \Delta t} \\
&= 2 \pi f(t_2, t_2)
\end{align*}
where the last equality follows from evaluating each term via the residue
theorem, closing the contour in the upper and lower half-plane
respectively. This requires $f(t_2+\Delta t, t_2)$ to grow more slowly
than $e^{\Omega \abs{\Im \Delta t}}$ over the arc at infinity, meaning $f$
is band-limited: for generic test functions $f$, we therefore take $\Omega
\to \infty$.Normalization convention
\begin{align*}
f(t) &= \int \dd{\omega} f(\omega) e^{-i \omega t} \\
f(\omega) &= \int \frac{\dd{t}}{2 \pi} f(t)e^{i \omega t} \\
\end{align*}
Space-time transform sign convention
\begin{align*}
f(\vb x, t) &= \int \dd[d]{\vb k}\dd{\omega} f(\vb k, \omega)e^{i \vb k \cdot \vb x - i \omega t} \\
f(\vb k, \omega) &= \int \frac{\dd[d]{\vb x}\dd{t}}{(2\pi)^{d+1}} f(\vb x, t)e^{-i \vb k \cdot \vb x + i \omega t} \\
\end{align*}
Explanation
The relative sign between $\vb k \cdot \vb x$ and $\omega$ is chosen to be "compatible" with the Schrodinger equation $i \pdv{f}{t} = -\pdv[2]{f}{x}$, yielding the de Broglie dispersion $\omega = k^2$. Ultimately the sign convention comes from the sign convention chosen for the Schrodinger equation.Sokhotski–Plemelj theorem
\begin{align*}
\int_0^\infty \dd{t} e^{-i \omega t} = \pi \delta(\omega) - \, \PV \frac{i}{\omega}
\end{align*}
Derivation
Identical to derivation of the delta function identity $\int \dd{t} e^{i \omega t} = 2 \pi \delta(\omega)$ except here the lower boundary term at $t = 0$ gives the integral against $1/\omega$. The indented contour used to evaluate the integral implies the Cauchy principal value. This can be brought to the usual form of Sokhotsi–Plemelj by explicitly parameterizing the shifted contour with $\omega \to \omega - i \epsilon$ where formally\[
\int_0^\infty \dd{t} e^{-i(\omega - i \epsilon)t} = \frac{-i}{\omega - i \epsilon}
\]