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.
Let $s=x(t-1)^n$. We then get $t = 1 + \left(\dfrac{s}x\right)^{1/n}$. We get $dt = \dfrac1{x^{1/n}} \dfrac{s^{1/n-1}}n ds$. Hence, $$I_n(x) = \int_1^2 \ln t \cdot e^{-x(t-1)^n}dt = \int_0^x \ln \left(1+\left(\dfrac{s}x \right)^{1/n}\right) e^{-s} \dfrac1{x^{1/n}} \dfrac{s^{1/n-1}}n ds$$ Hence, $$I_n(x) = \dfrac1{nx^{1/n}} \sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{x^{k/n}} \int_0^x s^{k/n} s^{1/n-1} e^{-s}ds$$ Hence, as $x \to \infty$ and keeping only the leading order term, we get $$I_n(x) \sim \dfrac1{nx^{1/n}} \dfrac1{x^{1/n}} \int_0^{\infty} s^{1/n} s^{1/n-1} e^{-s}ds = \dfrac1{nx^{2/n}}\int_0^{\infty} s^{2/n-1}e^{-s}ds$$