= Solution
<Watson's lemma> gives, under integrability away from zero and the stated endpoint <asymptotic expansion>,
$$
\boxed{I(\lambda)\sim\sum_{n=0}^\infty\frac{a_n\Gamma(n\beta+1)}{\lambda^{n\beta+1}}.}
$$
Indeed each power integrates over the positive half-line to $\Gamma(n\beta+1)\lambda^{-(n\beta+1)}$; replacing a fixed upper endpoint by infinity has exponentially small error. A remainder $o(t^{N\beta})$ near zero gives $o(\lambda^{-N\beta-1})$ after splitting off a small endpoint neighbourhood and rescaling $u=\lambda t$.
For <Laplace's method>, assume the phase and amplitude are smooth near the nondegenerate maximum, as required for a full expansion. Localize to a small neighbourhood of $c$: outside it the unique maximum and continuity make the exponential smaller by a factor $e^{-\lambda\varepsilon}$. Set
$$
\zeta=\operatorname{sgn}(t-c)\sqrt{2(\phi(c)-\phi(t))}.
$$
Near $c$ this is a smooth increasing coordinate, with $\zeta'(c)=\sqrt{|\phi''(c)|}$. It need not be monotonic on the entire integration interval. In the localized integral put $A(\zeta)=F(t(\zeta))\,dt/d\zeta$. Extending its Gaussian integration to both infinite endpoints incurs exponentially small error. Taylor expansion of $A$ and integration of even Gaussian moments gives
$$
J(\lambda)\sim e^{\lambda\phi(c)}\sqrt{2\pi}\sum_{j\geq0}
\frac{A^{(2j)}(0)}{2^j j!\,\lambda^{j+1/2}}.
$$
Here $A(0)=F(c)/\sqrt{|\phi''(c)|}$, giving \b[the leading term $\boxed{e^{\lambda\phi(c)}F(c)\sqrt{2\pi/(\lambda|\phi''(c)|)}}$]. The asymptotic-equivalence notation for that term presumes $F(c)\ne0$; if it vanishes the series determines the first nonzero term. The maximum condition alone, without amplitude regularity, would not imply a full expansion.
For the <gamma function>, set $t=xs$:
$$
\Gamma(x+1)=x^{x+1}\int_0^\infty e^{x(\log s-s)}\,ds.
$$
The phase has unique maximum $-1$ at $s=1$, with second derivative $-1$. To obtain the correction, put $s=1+u/\sqrt x$. Expansion gives
$$
x(\log s-s)=-x-\frac{u^2}{2}+\frac{u^3}{3\sqrt x}-\frac{u^4}{4x}+O(x^{-3/2}u^5),
$$
and hence a Gaussian multiplier
$$
1+\frac{u^3}{3\sqrt x}+\frac1x\left(-\frac{u^4}4+\frac{u^6}{18}\right)+\cdots.
$$
The odd term integrates to zero. The normalized Gaussian fourth and sixth moments are 3 and 15, so the correction is $-3/4+15/18=1/12$. Localization near one, with the exponentially suppressed endpoint tails, justifies termwise expansion. Thus <Stirling's formula> is
$$
\boxed{\Gamma(x+1)\sim\sqrt{2\pi}\,x^{x+1/2}e^{-x}\left(1+\frac1{12x}+O(x^{-2})\right).}
$$
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