State Watson's lemma giving an asymptotic expansion as $\lambda \rightarrow \infty$ for an integral of the form

$$I_{1}=\int_{0}^{A} f(t) e^{-\lambda t} d t, \quad A>0$$

Show how this result may be used to find an asymptotic expansion as $\lambda \rightarrow \infty$ for an integral of the form

$$I_{2}=\int_{-A}^{B} f(t) e^{-\lambda t^{2}} d t, \quad A>0, B>0$$

Hence derive Laplace's method for obtaining an asymptotic expansion as $\lambda \rightarrow \infty$ for an integral of the form

$$I_{3}=\int_{a}^{b} f(t) e^{\lambda \phi(t)} d t$$

where $\phi(t)$ is differentiable, for the cases: (i) $\phi^{\prime}(t)<0$ in $a \leq t \leq b$; and (ii) $\phi^{\prime}(t)$ has a simple zero at $t=c$ with $a<c<b$ and $\phi^{\prime \prime}(c)<0$.

Find the first two terms in the asymptotic expansion as $x \rightarrow \infty$ of

$$I_{4}=\int_{-\infty}^{\infty} \log \left(1+t^{2}\right) e^{-x t^{2}} d t$$

[You may leave your answer expressed in terms of $\Gamma$-functions.]