I am confused you speak of support of $f$ contained in $[-1/2,1/2]$ and $f=O(1/t^2)$.

For $\sigma>0$, I think the transform of $f(x):=\Gamma(\sigma+iax)$ is $$\widehat {f}(u)=\frac{2\pi}{a}\exp(2\pi \sigma u/a) \exp(-e^{2\pi u /a}).$$ Therefore, $\widehat{f}(u)\widehat{f}(-u)$ is the transform of a convolution of two gamma functions. (I am translating from my writing in 2006, this explain the parameters)

I see that the link to the paper by Landau do not give anything. So I put now the doi of the paper in Numdan: DOI: 10.24033/asens.595

@Ilya Bogdanov In your case the intersection $\{a\}$ is also non cancelling, because it is equal to $S_1\cap S_2$, $S_1\cap S_3$, $S_2\cap S_3$ and also to $S_1\cap S_2\cap S_3\}$ so its coefficient is $3(-1)+1(+1)=-2$. More explanation in my entry explaining the conjecture in the "Blog del Imus" institucional.us.es/blogimus/en/2024/03/still-hot-from-the-oven

It is increasingly common to see that the latest version of arXiv corrects errors in the journal version. I would look at the latest arXiv version of any journal published article.

The error is as $S(t)/\log t$. It is known that there are points $t$ as large as we want with $S(t)>c\sqrt{\log t/\log\log t}$. And for all $t$ we have $S(t)\le C \log t$ (I am assuming $t>10$, for example).

You can see also the case $b=0$ and $m=0$, the sum being $\log x+\gamma$. Starting at $j=1$.

When $b=-1$ and $m=0$ your sum is $$-e^x\int_x^\infty \frac{e^{- u}}{u}\,du$$ When $b=-1/2$ and $m=0$ your sum is $$-e^{x/2}K_0(x/2)$$ You can see these at V. Ditkine and A. Proudnikov, Calcul Operationnel

The references to the book are not very precise, which book is it about?