Use Watson's Lemma to find the asymptotic expansion of the following integral as $\lambda \to \infty$ with $\lambda>0.$ Assuming that $\phi (t)$ is infinitely differentiable on $[0,1].$ $$F(\lambda):=\int_{0}^{1}e^{-\lambda (2t-1)^4}\phi(t){d}t$$
My approach: Make successive substitutions: $s=t-\frac{1}{2}$, $\psi(s)=\phi(s+\frac{1}{2})$, $p=4s^2$ to get $$ F(\lambda) = \int_{-\frac{1}{2}}^{\frac{1}{2}}e^{-\lambda (2s)^4}\phi(s+\frac{1}{2}){d}s\\=\int_{-\frac{1}{2}}^{\frac{1}{2}}e^{-\lambda (2s)^4}\psi(s){d}s\\=2\int_{0}^{\frac{1}{2}}e^{-\lambda (2s)^4}\psi_{\mathrm{even}}(s){d}s \\=\frac{2}{4}\int_{0}^{1}e^{-\lambda p^2}p^{-\frac{1}{2}}\psi_{\mathrm{even}}(\frac{\sqrt{p}}{2}){d}p \\ $$
in which $\psi_{\mathrm{even}}(s):=\frac{\psi(s)+\psi(-s)}{2}$ is infinitely differentiable on $[-\frac{1}{2},\frac{1}{2}].$
It turns out that the $e^{-\lambda p^2}$ part in the last integral still prevents a direct application of Watson's Lemma?