In a response to one of my earlier questions which I believe was related to Evolution of Zeta Zeros from Fourier Transform of $e^{-t/2}\left(\psi'[e^t]-1\right)$, it was suggested I instead focus on the Fourier transform of $\frac{\psi[e^u]-e^u}{e^{u(1/2+\epsilon)}}$ where $\psi[x]$ is the second Chebyshev function. Evaluation of this Fourier transform results in terms which contain Dirac delta functions with complex arguments such as the following.
$$\text{FourierTransform}\left[\frac{-e^u}{e^{u \left(\frac{1}{2}+\epsilon \right)}},u,y\right]=-2\ \sqrt{2\ \pi}\ \delta\ [-i+2\ y+2\ i\ \epsilon]$$
Question 1: What is the meaning of a Dirac delta function with a complex argument? In what direction does the integral of a Dirac delta function with a complex argument evaluate to a non-zero result?
Evaluation of the suggested Fourier transform using the Fourier series representation of $\psi[x]$ (see Illustration of Fourier Series for Prime Counting Functions) results in additional terms which contain Dirac delta functions with complex arguments. I believe I can get some, but not all, of these terms which contain Dirac delta functions with complex arguments to cancel each other out.
Question 2: Since the Fourier transform is being evaluated over $y\in Reals$, can I assume all terms containing Dirac delta functions with complex arguments can safely be ignored?
Everything is explained in the books on the Riemann zeta function (Titchmarsh, Apostol, Edwards, Montgomery, Iwaniec..). And I don't understand what you did, so in short there is what you need to know :
From $\frac{\zeta'(s)}{\zeta(s)} = -s\int_1^\infty \psi(x)x^{-s-1}dx$, by inverse Mellin transform and the residue theorem, we have the Riemann explicit formula $$\psi(x) = \dfrac{1}{2\pi i}\int_{\sigma-i \infty}^{\sigma-i \infty}\left(-\dfrac{\zeta'(s)}{\zeta(s)}\right)\dfrac{x^s}{s}ds=x-\sum_\rho\frac{x^\rho}{\rho} - \log(2\pi) -\dfrac{1}{2}\log(1-x^{-2})$$
If the RH is true, with $\rho = 1/2+i\lambda$ it means that $$f(u) = \frac{\psi(e^u)-e^u+ \log(2\pi)+\dfrac{1}{2}\log(1-e^{-2u})}{e^{u/2}} = -\sum_\lambda \frac{e^{i \lambda u}}{1/2+i\lambda} = \int_{-\infty}^\infty \hat{f}(\omega)e^{i \omega u}d\omega$$ where $\hat{f}(\omega) = -\sum_\lambda \frac{\delta(\omega-\lambda)}{1/2+i\lambda} $ is the Fourier transform (in the sense of distributions) of $f(u)$.
Note that both $f'(u)$ and its Fourier transform is (up to a very simple term) a series of Dirac deltas, those kind of distributions being very rare (lattices, crystals, quasicrystals)