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.