How do I find the first term in the asymptotic expansion of the following.
$$\int_0^1 e^{-x(t^4+1)}\sin(t) \ dt$$ as $x \rightarrow \infty$.
Attempt:
First replace $\sin(t)=t$ (Taylor series expansion - only need first term).
Therefore we have $e^{x}\int_0^1 e^{-xt^4}t \ dt $.
I am not sure on how to finish the steps - do we replace $e^{-xt^4}$ by it's Taylor series as well and then integrate term by term?
$$ \begin{align} \int_0^1e^{-x(t^4+1)}\sin(t)\mathrm{d}t &=e^{-x}\int_0^1e^{-xt^4}\sin(t)\mathrm{d}t\\ &=e^{-x}\int_0^1e^{-xt^4}\left(t+O\!\left(t^3\right)\right)\mathrm{d}t\\ &=\frac14e^{-x}\int_0^1e^{-xt}\left(t^{-1/2}+O\!\left(1\right)\right)\mathrm{d}t\\ &=\frac14e^{-x}\int_0^xe^{-t}\left(x^{-1/2}t^{-1/2}+x^{-1}O(1)\right)\mathrm{d}t\\ &=\frac14x^{-1/2}e^{-x}\int_0^xe^{-t}t^{-1/2}\mathrm{d}t+\frac14x^{-1}e^{-x}\int_0^xe^{-t}O(1)\mathrm{d}t\\ &=\frac{\sqrt\pi}4x^{-1/2}e^{-x}+O\!\left(x^{-1}e^{-x}\right)\\ \end{align} $$