the maximum possible value of $Q$ in we get $$\left\vert {\operatorname{Im}}z\right\vert <\frac{9\sqrt{2}}{8}c\sqrt{t}$$and of Theorem \[thm1.2\] follows. Thus, Theorem \[thm1.2\] is proven.

-   $0<\varepsilon <1/2$. Then $\frac{\varepsilon }{1-\varepsilon }<1$ and clearly diverges for any ${\operatorname{Im}}z\neq 0$ and our method fails to establish analyticity and we have to go back to and analyze it for any natural $n$. Expanding $\left( \left\vert \alpha \right\vert +\beta \right) ^{n-1}$ in by the binomial formula we have $$\begin{gathered}
      \label{eq5.6.1}
    \int_{0}^{\varkappa /2}\frac{d\beta }{\beta }\int_{-\infty }^{\infty
    }F(\alpha ,\beta )d\alpha \\
    \lesssim _{z,t,q_{+}}\sum_{k=0}^{n-1}\binom{n-1}{k}\int_{0}^{\varkappa
    /2}d\beta \beta ^{k-1}e^{-Q\beta ^{-\frac{\varepsilon }{1-\varepsilon }}}\int_{0}^{\infty }\alpha ^{n-k-1}e^{-24\beta t\alpha ^{2}}d\alpha .\end{gathered}$$Reducing the inner integral in to the gamma function[^8], $$\begin{aligned}
    \eqref{eq5.6.1}& =\sum_{k=0}^{n-1}\binom{n-1}{k}\int_{0}^{\varkappa
    /2}d\beta \beta ^{k-1}e^{-Q\beta ^{-\frac{\varepsilon }{1-\varepsilon }}}\cdot \frac{1}{(3\beta t)^{\frac{n-k}{2}}}\Gamma \left( \frac{n-k}{2}\right)  \notag \\
    & \lesssim \sum_{k=0}^{n}\binom{n}{k}(3t)^{-\frac{n-k}{2}}\Gamma \left( 
    \frac{n-k}{2}\right) \int_{0}^{1}d\beta \beta ^{\frac{3k-n}{2}-1}e^{-Q\beta
    ^{-\frac{\varepsilon }{1-\varepsilon }}}.  \label{eq5.7.1}\end{aligned}$$Introducing in the last integral the new variable $s=\beta ^{-\frac{\varepsilon }{1-\varepsilon }}$ and setting $\gamma \equiv\frac{1-\varepsilon }{2\varepsilon }>1/2$ we get $$\begin{aligned}
    \int_{0}^{1}d\beta \beta ^{\frac{3k-n}{2}-1}e^{-Q\beta ^{-\frac{\varepsilon 
    }{1-\varepsilon }}}& =\frac{1-\varepsilon }{\varepsilon }\int_{1}^{\infty
    }s^{\frac{\varepsilon -1}{\varepsilon }\left( \frac{3k-n}{2}-1\right)
    -1}e^{-Qs}ds  \notag \\
    & \lesssim \int_{1}^{\infty }s^{\gamma (n-3k)-1}e^{-Qs}ds  \notag \\
    & \lesssim Q^{-\gamma (n-3k)+1}\int_{Q}^{\infty }s^{\gamma (n-3k)}e^{-s}ds.
    \label{eq5.7.2}\end{aligned}$$The behavior of the last integral depends on the sign of $\omega
    _{k}\equiv\gamma (n-3k)$. If $\omega _{k}\geq 0$, i.e. $3k\leq n$, 