Consider the integral

\begin{equation}

I_n(x)=\int^{2}_{1} (log_{e}t) e^{-x(t-1)^{n}}dt

\end{equation}

Use Laplace's Method to show that

\begin{equation}

I_n(x) \sim \frac{1}{nx^\frac{2}{n}} \int^{\infty}_{0} \tau^{\frac{2-n}{n}} e^{-\tau} d\tau \end{equation}

as $x\rightarrow\infty$.

where $0<n\leq2$. Hence find the leading order behaviour of $I_{1}(x)$. and $I_{2}(x)$ as $x\rightarrow \infty$.

=>

Its really difficult question for me.

Here,

$g(t) = -(t-1)^{n}$ has the maximum at $t=0$

but $h(t)= log_{e}t$ at $t=0$

$h(0)=0$.

so I can not go any further. PLEASE HELP ME.