Asymptotics of Laplace like integral with shrinking integration intervals I want to find the asymptotics of the integral of the form
$$I(M) = \int_0^1 x^M e^{-M f(\frac{x}{\ln M})} dx$$
as $M \to +\infty$. Assume also that $f$ is analytic and increasing near $0$, with power series expansion:
$$f(x) = \frac{1}{2} x^2 + c_1 x^3 + c_2 x^4 +...$$
and all coefficients $c_n \geq 0$ (we can put additional assumptions on $f$ if necessary). In particular $f$ attains its minimum at $0$ and also $f'(0)=0$.
I'm not sure how to proceed as it is not clear what is the dominant behavior of the integrand because for any $n\geq 2$
$$\frac{M}{(\ln M)^n} \to +\infty, \qquad as \ M \to +\infty$$
So we cannot stop at a certain term in the series expansion of $f$ and obtain the dominant term of the integrand. We can also change the variables
$$I(M) = (\ln M)^{M+1} \int_0^{\frac{1}{\ln M}} x^M e^{-Mf(x)} dx$$
 A: With the change of variables $x=e^{-t}$, we have
$$
I(M) = (\log M)^{M + 1} \int_{\log \log M}^{ + \infty } {e^{ - Mg(t)} e^{ - t} dt} ,
$$
where $g(t) = t+ f(e^{ - t} )$. This further equals to
\begin{align*}
I(M) &= (\log M)^{M + 1} e^{ - Mg(\log \log M)} \int_{\log \log M}^{ + \infty } {e^{ - M(g(t) - g(\log \log M))} e^{ - t} dt} \\ & = (\log M)\exp \left[ { - Mf\left( {\frac{1}{{\log M}}} \right)} \right]\int_{\log \log M}^{ + \infty } {e^{ - M(g(t) - g(\log \log M))} e^{ - t} dt} .
\end{align*}
Now $g'(t) = 1 - f'(e^{ - t} )e^{ - t}  = 1 + \mathcal{O}(e^{ - 2t} ) > 0$ for large $t$, i.e., the function $g(t)$ is increasing for sufficiently large $t$. Thus, there is a linear dependence at the endpoint $t=\log\log M$ (i.e., it is not a global minimum point of $g(t)$) for all large $M$. Hence, by the general Laplace method,
\begin{align*}
\int_{\log \log M}^{ + \infty } {e^{ - M(g(t) - g(\log \log M))} e^{ - t} dt} & \sim \frac{1}{{M\log M}}\frac{1}{{g'(\log \log M)}}
\\ & =\frac{1}{{M\log M}}\frac{1}{{1 - f'\!\left( {\frac{1}{{\log M}}} \right)\frac{1}{{\log M}}}} \\ &\sim \frac{1}{{M\log M}}
\end{align*}
as $M\to+\infty$. The last asymptotic equality follows from the fact that $f'\!\left( {\frac{1}{{\log M}}} \right)\frac{1}{{\log M}} = \mathcal{O}\!\left( {\frac{1}{{\log ^2 M}}} \right)$, which in turn follows from the series expansion of $f(x)$. Accordingly,
$$
I(M) \sim \frac{1}{M}\exp \left[ { - Mf\left( {\frac{1}{{\log M}}} \right)} \right]
$$
as $M\to +\infty$.
