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

=>
Its really difficult question for me.

Here,

[itex]g(t) = -(t-1)^{n}[/itex] has the maximum at [itex]t=0[/itex]

but [itex]h(t)= log_{e}t[/itex] at [itex]t=0[/itex]
[itex]h(0)=0[/itex].
so I can not go any further. PLEASE HELP ME.

 

[Xiuh]

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

 

[wel]

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

at [itex] t=1 [/itex], [itex] g(t) =0 [/itex], how can i say it is maximum?

what is the Taylor expansion of the [itex] log_{e} t[/itex]?

 

[Xiuh]

at [itex] t=1 [/itex], [itex] g(t) =0 [/itex], how can i say it is maximum?

what is the Taylor expansion of the [itex] log_{e} t[/itex]?

[itex] g [/itex] is decreasing.

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