MHB How Can Laplace's Method Simplify This Integral?

[ra_forever8]

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.

 

[chisigma]

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.

First it is useful to write the integral in a more suitable form...

$\displaystyle I_{n} (x) = \int_{0}^{1} \ln (1 + t)\ e^{- x\ t^{n}}\ d t\ (9)$

The Laplace integration of an expression like...

$\displaystyle I(x) = \int_{a}^{b} f(t)\ e^{- x\ g(t)}\ dt\ (2)$

... consists in approximating (2) with an expression which is valid for 'large' x according with the following steps...

a) let's suppose that g(t) has a minimum in t= c, being c in [a,b]...

b) we expand in Taylor series f(t) and g(t) around t=c...

c) we extend the integration range in $(- \infty, + \infty)$ or in $[0, + \infty)$ if c=0 and try to use the properties of the Laplace Transform [that's why this procedure is called 'Laplace Integration'...]

Now we can procced taking into account that is $\displaystyle f(t) = \ln (1 + t) \sim t$, $\displaystyle g(t)= t^{n}$ which has a minimum in x=0...

$\displaystyle I_{n} (x) = \int_{0}^{1} \ln (1 + t)\ e^{- x\ t^{n}}\ d t \sim \frac{1}{n}\ \int_{0}^{\infty} \tau^{\frac{1}{n}-1}\ \ln (1 + \tau^{\frac{1}{n}})\ e^{- x\ \tau}\ d \tau\ \sim$

$\displaystyle \sim \frac{1}{n}\ \int_{0}^{\infty} \tau^{\frac{2}{n}-1}\ e^{- x\ \tau}\ d \tau = \frac{\Gamma (\frac{2}{n})}{n\ x^{\frac{2}{n}}}\ (3)$

Kind regards

$\chi$ $\sigma$