Consider $$L_n = \frac{1}{2\pi i } \oint_{C'} \frac{1}{(1-t)^{\alpha+1} t^{n+1}} e^{-\frac{xt}{1-t}} dt\,\,\,\,(1)$$ where $C'$ is an anticlockwise contour around zero. Now set $\alpha = n$ and I want to find the leading asymptotic behaviour for large $n$.
Attempt: Use the saddle point approximation, so need to put integral in form $$L_n = \int_C g(t) e^{n f(t)} dt$$ I thought about reexpressing the terms dependent on $n$ in the integrand $(1)$ there by exponentials so I could use the method but my choice results in a vanishing second derivative of the function multiplying $n$.
The integral can be rewritten in the following way $$\frac{1}{2 \pi i} \oint_{C'} \exp(n(\underbrace{-\ln(t(1-t)))}_{f(t)}) \underbrace{\exp(-\ln(t(1-t))) \exp(-xt/1-t)}_{g(t)} dt$$
Then $g(t) \approx g(t_o)$ and $f(t) \approx f(t_o) + f''(t_o) (t-t_o)^2/2$. My problem is when I evaluate $f''$ it vanishes. Should I expand to next power? Many thanks.