Asymptotic approximation of an integral.

Question

How to find the asymtotic approximation of this Fourier integral : $$I[T]=\lim_{T->\infty}^{}\frac{T}{2\pi}\int_{-\pi/T}^{\pi/T}d\lambda e^{iT\lambda}e^{Tf(\lambda)}.$$ $\textbf{Context}$ : This is a simplification of the problem i have encountered in the context of statistical analysis of a physics problem (specifically counting statistics of particles flux between two reservoirs), where moment generating function for particle flux is given by as : $$Z[\chi,T]=\frac{T}{2\pi}\int_{-\pi/T}^{\pi/T}d\lambda e^{iT\lambda}e^{Tf(\lambda,\chi)},$$ with $f(\lambda,\chi)$ being the unconstrained (here $\lambda$ integral is used to impose a constraint) scaled cumulant generating function. Specifically I am interested in long time limit of the scaled cumulant generating function , defined as : $$\lim_{T->\infty}=\frac{1}{T}\ln\big(Z[\chi,T]\big),$$ whether it is independent of $T$. This can be thought of as getting a long time limit moment generating function for a contracted (one variable being fixed) bi-variate probability distribution.

Answer 1

Proceeding very naively, and probably making errors along the way, I get that the limit does not exist if $f(0) > 0$, is zero if $f(0) < 0$, and depends on $f'(0)$ if $f(0) = 0$.

Here is my fiddling, replacing $T$ by $t$ and $\lambda$ by $x$.

$\begin{array}\\ I &=\lim_{t->\infty}\frac{t}{2\pi}\int_{-\pi/t}^{\pi/t}dx e^{itx}e^{tf(x)}\\ &=\lim_{t->\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}dy e^{iy}e^{tf(y/t)} \qquad x = y/t\\ &\approx\lim_{t->\infty}\frac{1}{2\pi}\int_{-\pi}^{\pi}dy e^{iy}e^{t(f(0)+(y/t)f'(0))} \qquad\text{since } y/t \text{ is small}\\ &=\lim_{t->\infty}\frac{e^{tf(0)}}{2\pi}\int_{-\pi}^{\pi}dy e^{y(i+f'(0))}\\ &=\lim_{t->\infty}\frac{e^{tf(0)}}{2\pi}\dfrac{e^{y(i+f'(0))}}{i+f'(0)}|_{-\pi}^{\pi}\\ &=\lim_{t->\infty}\frac{e^{tf(0)}}{2\pi}\dfrac{e^{\pi(i+f'(0))}-e^{-\pi(i+f'(0))}}{i+f'(0)}\\ &=\lim_{t->\infty}\frac{e^{tf(0)}}{2\pi}\dfrac{-e^{\pi f'(0)}+e^{-\pi f'(0)}}{i+f'(0)}\\ \end{array} $