State Watson's lemma, describing the asymptotic behaviour of the integral

$$I(\lambda)=\int_{0}^{A} e^{-\lambda t} f(t) d t, \quad A>0$$

as $\lambda \rightarrow \infty$, given that $f(t)$ has the asymptotic expansion

$$f(t) \sim \sum_{n=0}^{\infty} a_{n} t^{n \beta}$$

as $t \rightarrow 0_{+}$, where $\beta>0$.

Consider the integral

$$J(\lambda)=\int_{a}^{b} e^{\lambda \phi(t)} F(t) d t,$$

where $\lambda \gg 1$ and $\phi(t)$ has a unique maximum in the interval $[a, b]$ at $c$, with $a<c<b$, such that

$$\phi^{\prime}(c)=0, \quad \phi^{\prime \prime}(c)<0 .$$

By using the change of variable from $t$ to $\zeta$, defined by

$$\phi(t)-\phi(c)=-\zeta^{2}$$

deduce an asymptotic expansion for $J(\lambda)$ as $\lambda \rightarrow \infty$. Show that the leading-order term gives

$$J(\lambda) \sim e^{\lambda \phi(c)} F(c)\left(\frac{2 \pi}{\lambda\left|\phi^{\prime \prime}(c)\right|}\right)^{\frac{1}{2}}$$

The gamma function $\Gamma(x)$ is defined for $x>0$ by

$$\Gamma(x)=\int_{0}^{\infty} e^{(x-1) \log t-t} d t$$

By means of the substitution $t=(x-1) s$, or otherwise, deduce that

$$\Gamma(x+1) \sim x^{\left(x+\frac{1}{2}\right)} e^{-x} \sqrt{2 \pi}\left(1+\frac{1}{12 x}+\ldots\right)$$

as $x \rightarrow \infty$