The coefficients for the Maclaurin series of the reciprocal $\Gamma$ function can be calculated by an integral formula given by: $$\frac{1}{\Gamma(z)}= \Sigma_{n} a_n z^n, \;a_n =\frac{(-1)^n}{\pi n!} \int_0^\infty e^{-t} \Im{[(\log(x)-i\pi)^n]} dt$$ as shown by Fekih-Ahmed (2014) in https://arxiv.org/abs/1407.5983. This integral formula yields a remarkable simple asymptotic expression: $$ a_n \simeq \frac{(-1)^n}{n!} \sqrt{\frac{2 n}{ \pi}} \Im{\left[\frac{z_0^{1/2-n} e^{-nz_0}}{\sqrt{1+z_0}}\right]}, \; z_0 = 1/W_{-1}(-n)$$ where $W_{-1}$ is the -1 branch of the Lambert $W$ function, through use of the saddle point approximation as derived by Fekih-Ahmed in the same paper.
In another related problem, I have come to a formula with similar expression: $$I_{n,k} = \int_0^1 (1-x)^n \Im{[(\log(x)-i\pi)^k]} dx$$ but even though I employ the same approach of the paper, my saddle point approximation seems to diverge from the exact integral as $k \rightarrow \infty$ for any value of $n$.
The general idea is to use saddle point approximation for the following integral: $$ I_{n,k} = \int_0^{1} e^{k[\log(1-z)/r+\log(\log(z)-i\pi)]} dz$$ so we seek to minimize $f(z) = \log(1-z)/r+\log(\log(z)-i\pi)$, where $r = k/n$, for $k \geq n$. (For now I am interested in fixing $n$ and increasing $k$.) We can find an analytical expression for the saddle point: $z_0 = 1/W_{-1}(-re^r)$, but while for small $k$ the formula like the one above yields results 10% different from the exact, it increases to 35% for larger $k$s. I was hoping to find an approximation with accuracy that oscillates but has no degrading trend. The approximation becomes quite awful as we increase the value of $n$. So I wonder if I am not applying the saddle point approximation correctly or if anyone has any better idea to approximate the value of $I_{n,k}$. I think the leading term should asymptotically become $\log^k(n)$.
Thank you!