Question: For $z \gg 1$, find the leading-order approximation to the integral, $$\int_0^{\infty} e^{t-z(t^4-2t^2)}\sin^2(2\pi\nu t)~dt,$$ allowing for any value of the parameter $\nu > 0$.
My approach: We first rewrite the integral as a Laplace integral: $$I(z) = \int_0^{\infty}f(t)e^{z\phi(t)}dt = \int_0^{\infty}e^{z[2t^2-t^4)]}e^t\sin^2(2\pi\nu t)~dt.$$ Therefore, $\phi(t) = 2t^2-t^4$ and $f(t) = e^t\sin^2(2\pi\nu t)$. It is easy to see that $\phi(t)$ attains its only maxima at $t=1$, which is an interior point. Now, from Laplace's method, we know that the leading order approximation of the integral $$J(z) = \int_a^b f(t)e^{z\phi(t)}~dt,$$ given that $\phi(t)$ attain its only maxima at an interior point $t=c$ ($a<c<b$), is $$J(z)\sim \frac{f(c)e^{x\phi(c)}}{\sqrt{x}}\int_{-\infty}^{\infty}e^{s^2\phi''(c)/2}ds = \frac{\sqrt{2\pi} f(c)e^{x\phi(c)}}{\sqrt{-x\phi''(c)}}.$$ Therefore, the leading order approximation of $I(z)$ is as follows: $$I(z) \sim \frac{\sqrt{2\pi}e\sin^2(2\pi\nu)e^x}{\sqrt{8x}}=\frac{\sqrt{\pi}e}{2}\sin^2(2\pi\nu)\frac{e^x}{\sqrt{x}}.$$
Can someone please let me know if this is the correct way to solve this problem. Also, please let me know if there's any other way to solve this problem. Thanks in advance!
Your solution is correct for the dominating term. A natural continuation is to see what the next terms are for the asymptotic expansion. I adopt the notation from DLMF. Let
$$ I(z) = \int_0^{\infty} e^{-zp(t)} q(t) dt $$
where $p(t)=t^4-2t^2, q(t)=e^t \sin^2{\xi t}$ ($\xi = 2\pi \nu$). An asymptotic expansion to the integral may be written as
$$ \int_a^b e^{-zp(t)}q(t)dt \sim e^{-zp(a)} \sum_{s=0}^{\infty} \Gamma\left(\frac{s+\lambda}{\mu} \right) \frac{b_s}{z^{(s+\lambda)/\mu}} $$
when a mimimum occurs at the lower limit $a$ and the constants depend upon $p(t)$ and $q(t)$.
As you have identified the minimum of $p(t)$ at $t=1$, we split the integral into two parts, from $0$ to $1$ ($I_1$) and from $1$ to $\infty$ ($I_2$).
For $I_2$, the minimum of $p(t)$ occurs at $a=1$ around which we have the series
\begin{align} p(t) &= -1 + 4(t-1)^2+4(t-1)^3 +(t-1)^4 \\ q(t) &= e \sin^2 \xi + e\sin\xi(\sin\xi+2\xi\sin\xi)(t-1)+... \end{align}
We have (from the link) $p(1)=-1, p_0=4,\mu=2$ and $q_0=e\sin^2\xi, \lambda=1$.
The three first terms are given explicitly in the link, one may also calculate the residues to obtain the coefficients $b_s$ from
\begin{align} b_s &= \frac{1}{\mu} \text{res}_{t=a} \left[\frac{q(t)}{(p(t)-p(a))^{(\lambda+s)/\mu}} \right] \\ &= \frac{1}{2}\text{res}_{t=1} \left[\frac{e^t \sin^2(\xi t)}{((t-1)^2(t+1)^2)^{(s+\lambda)/\mu}} \right] \\ \end{align}
The three first coefficients are
\begin{align} b_0 &= \frac{1}{4}e \sin^2\xi \\ b_1 &= -\frac{1}{4}e\xi\sin\xi\cos\xi \\ b_2 &= \frac{1}{64}e((4\xi^2-1)\cos2\xi-2\xi\sin2\xi+1) \end{align}
We do the same for $I_1$ and note that the coefficient $b_0$ is the same for both intervals. However, the second coefficient $b_1$ cancels whereas $b_2$ is the same for both intervals. We obtain
$$ \int_0^{\infty} e^{-z(t^4-2t^2)} e^t \sin^2{\xi t}dt \sim \frac{\sqrt{\pi}}{2} z^{-1/2}e^{z+1}\left[\sin^2{\xi} + \frac{z^{-1}}{32}((4\xi^2-1)\cos2\xi-2\xi\sin 2\xi+1)\right] $$
Note that the first term is same as yours. Also note, when $\xi$ increases, the expression eventually breaks down.