Asymptotic Expansion of Integrals Using Laplace's Method

[wel]

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.

 

[Xiuh]

I believe you meant $g$ reaches its maximum at $t = 1$.
Indeed, $h(1) = 0$, but that shouldn't be a problem, just use higher order terms in the Taylor expansion of the logarithm.

 

[wel]

I believe you meant $g$ reaches its maximum at $t = 1$.
Indeed, $h(1) = 0$, but that shouldn't be a problem, just use higher order terms in the Taylor expansion of the logarithm.

at $t=1$, $g(t) =0$, how can i say it is maximum?

what is the Taylor expansion of the $log_{e} t$?

 

[Xiuh]

at $t=1$, $g(t) =0$, how can i say it is maximum?

what is the Taylor expansion of the $log_{e} t$?

$g$ is decreasing.

Do you remember what a Taylor series is? That's basic if you want to calculate asymptotic expansions of integrals.