I genuinely appreciate your patience. The question is simple now; The integrability of $x\mapsto \psi(x)/x$ is sufficient for the asymptotic $(12)$. Is it necessary ? (hence the numerical experiments...)

Thank you so much. The assumption that $\psi(x)/x$ is integrable is strong. Numerical experiments suggest that $I(t)\sim \log{t}/\sqrt{t}$ when tested with $\psi$ a positive bump centered at $x=0$ or a Gaussian. I think we should only need $\psi(x)/x$ to be locally integrable away from zero, which is the case of course.

Now, it remains to remove the assumption that $\psi(0)\ne 0$. I appreciate your effort and I really want to accept your answer but I fear the question will no longer be of interest to others. So (+1) for now.

The statement ''So, letting $A$ go to $\infty$ slowly enough, we will have $A=o(\sqrt t)$'' is not really necessary I think. One can take $A=1$ and get a precise estimate for $\int_0^1=o(1/\sqrt{t})$.

A smooth bump. No smallness near zero is given if that is your question.

@Iosif Pinelis You are right of course. But the context is $\phi^{\prime\prime}<c<0$ in the support of a smooth cut-off function. The latter has compact support small enough that contains only one local maximum point of $\phi$. Near its local maximum, $\phi$ is concave.

When $\phi(x_{0})>0$ and $\phi^{\prime\prime}<0$, one can look at a small enough neighborhood of $x_{0}$. But in this case $\phi(x)$ is merely bounded below by $\phi(x_{0})$. This finishes the argument. The statement $\phi(x)\gtrsim (x-x_{0})^2$ is not true however when $\phi^{\prime\prime}<0$ as shown by the counterexample $\phi(x)=1-x^2$ with $x_{0}=0$. Thank you for the useful comments.

Thank you Christian. I don't know how to show $\phi(x)\gtrsim (x-x_{0})^2$ when $\phi^{\prime\prime}<c<0$. This is the last missing piece indeed. Could you make it rigorous ?