In this MO question here, I asked about the Fourier transform of the zeta function. The second answer lists the following as a representation for $\zeta(s)$, with $E(x)$ as the floor function:
\begin{multline} \zeta(\frac 1 2 +it)=-it\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 -it }dx -\frac 1 2\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 -it }dx\\-\frac{1+2it}{1-2it}, \tag{$\ast$} \end{multline}
The idea is that by getting the Fourier transform of the expression $\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 -it }dx$, we can essentially get the whole thing.
To get this Fourier transform, we need to solve the double integral
$F(\omega) = \int_{-\infty}^{\infty} \left[\int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 -it }dx \right] e^{-i\omega t} dt$
The answer then goes onto say:
With the above formula, it is easy to find an explicit expression for the Fourier transform: in fact, we need only to calculate the Fourier transform of $t\mapsto e^{-it \ln x}$, which is $\delta_0(\tau+\frac{\ln x}{2π})$... As a result, the Fourier transform of the second term in $(\ast)$ is given by $$ \int_{1}^{+\infty}\bigl(x-E(x)\bigr)x^{-\frac3 2 }\delta_0(\tau+\frac{\ln x}{2π})dx, $$
I am somewhat at a loss for how to arrive at this result. How do you go from the double integral to this? It doesn't seem like you can flip the order of the integrals here since there's that $e^{-i\omega t}$ on the outside.