Consider the integral

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

in the limit $\lambda \rightarrow \infty$, 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 $\beta>0$. State Watson's lemma.

Now consider the integral

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

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

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

By making a monotonic change of variable from $t$ to a suitable variable $\zeta$ (Laplace's method), or otherwise, deduce the existence of an asymptotic expansion for $J(\lambda)$ as $\lambda \rightarrow \infty$. Derive the leading term

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

The gamma function is defined for $x>0$ by

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

By means of the substitution $t=x s$, or otherwise, deduce Stirling's formula

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

as $x \rightarrow \infty$