in the Supplement of \[18; Sect. 3\]. So, first we represent $F_{m,b}(e^{-\alpha_n})$ using eq. (\[efem\]) of Lemma 8 and then, we apply Perron formula to the partial sum $D_m(s)$ of the Dirichlet series $D(s)$ (recall also (\[dmofes\]) and (\[dofes\])). In this way we arrive at the following complex integral representation: for any $\Delta>1$, we have $$\begin{aligned}
 \label{doubleint}
& & F_{m,b}(e^{-\alpha_n})  \\
& & = \frac{1}{2\pi i}
\int_{\rho+\Delta-i\infty}^{\rho+\Delta+i\infty}
\alpha_n^{-s}\Gamma(s)\zeta(s+1)
\left(\frac{A(m+1)^{\rho-s}}{\rho-s} +D(s) +\Omega_m\right) ds,
\nonumber\end{aligned}$$ where $\Omega_m=o(1), m\to\infty$. Furthermore, (\[dfas\]) and (\[doubleint\]) imply that $$\begin{aligned}
 \label{dfasint}
& & \mathbb{P}(X_n\le m)  \\
& & \sim \exp{\left\{-\frac{A\alpha_n^{-\rho}}{2\pi i}
\int_{\Delta-i\infty}^{\Delta+i\infty} ((m+1)\alpha_n)^{-s}
\Gamma(s+\rho) \zeta(s+\rho+1) \frac{ds}{s}\right\}}. \nonumber\end{aligned}$$ The proofs of (\[doubleint\]) and (\[dfasint\]) contain some technical details that will be given in the Appendix.

We continue with the computation of the complex integral in the exponent of (\[dfasint\]). The sequence $m=m(n)$ will be specified later in a more precise way. At this moment we only asume that it satisfies (\[mn\]). We set in the integral of (\[dfasint\]) $$\label{un}
u=u_n=(m+1)\alpha_n,$$ and consider it as a function of $u$. First, we shall obtain its explicit form and then we shall estimate it as $u\to\infty$ (see (\[mn\]) and (\[un\])). Clearly, we can consider this integral as the inverse Mellin transform 