$ I(n) =\int^\infty_0e^{-\lambda t-n\ln t}\int^\infty_{-\infty}e^{-s^2/2}e^{-i(s/\sigma)\ln t}e^{i(s/\sigma)\ln n}/\Gamma(1+is/\sigma) dsdt $
as $n\rightarrow \infty$, $\lambda >0$ . This is what I did.
Change variables $t=nz$, so $dt=ndz$. Thus $\ln t=\ln n +\ln z$ and \begin{align} &e^{- \lambda t-n \ln( t)} =n^{-n}e^{-n(\lambda z+\ln z)}=n^{-n}e^{-n\xi(z)}\\ e^{-(is/\sigma)\ln t} &=e^{-(is/\sigma)\ln n}e^{-(is/\sigma)\ln z} \end{align} where $\xi(z)=\lambda z+\ln z$. With this transformation, \begin{align} I(n) &=nn^{-n}\int^\infty_0e^{-n\xi(z)}h(z)dz\\ h(z)= &\int^\infty_{-\infty}e^{-s^2/2}e^{-i(s/\sigma)\ln z}/\Gamma(1+is/\sigma)ds \end{align}
Thusit is in the form for Laplace's approximation. The problem is, the critical point $z_0=-1/\lambda$ of $\xi(z)$ does not lie in $(0,\infty)$.
So, I don't think this is the right approach. Any ideas, I welcome.