Find the leading order behaviour of $$I(x)=\int_0^1 \sin te^{-x\sinh^4 t}dt, \enspace x\rightarrow\infty$$ Thus evaluate $I(10)$ and compare it with its numerical value $I(10)\approx0.13564$
So far, I have defined $f(t)=\sin t$ and $g(t)=-\sinh^4 t$ so that I can use Laplace's Method. Then, to find where $g(t)_{\max}$ occurs, $g'(t)=-4\sinh^3 t\cosh t=0$ and so $t=0$. (Also it should be noted that we have been introduced to Laplace's Method as an application of Watson's Lemma only, where we can disregard the non-local contributions as we have proved they are non-dominant.)
I have been given the following fact that I can use:
If $g(t)$ achieves its maximum value $g(t)_{\max}=\max_{a \leq t\leq b}g(t)$ at a set of points in the interval of integration $[a,b]$, then let $c_i$ denote the points where the maximum is achieved (for $i=1,…,N)$. Laplace's method shows that if $g''(c_i)<0$ for all $i$, then \begin{equation}
\begin{aligned}
I(x) &= \int_a^be^{xg(t)}f(t)dt \\
& = e^{xg_{\max}} \left( x^{-\frac{1}{2}}\sum_{i=1}^N \sqrt \frac{2\pi}{-g''(c_i)}f(c_i)+O(x^{-1}) \right) \enspace x\rightarrow\infty
\end{aligned}
\end{equation} Where $O(x^{-1})$ represents the Big-Oh notation.
Seeing as I know that $c=0$, when I substitute this $c$ value into $g''(t)=-4\sinh^4t-12\sinh^2t\cosh^2t$, I get $g''(0)=0$. However, this throws up issues since using the rule above, I would have $g''(c=0)=0$ as a denominator.
Have I misinterpreted the question, or is there a better method?
Best Answer
You can make the change of variables $t = \operatorname{arcsinh} \sqrt u$ to reduce the integral to the standard case. Essentially, this holds: $$\int_0^1 e^{-x \sinh^4 t} \sin t\,dt \sim \int_0^\infty e^{-x t^4} t \,dt = \frac 1 2 \int_0^\infty e^{-x (t^2)^2} d(t^2).$$