Two real functions $p(t), q(t)$ of a real variable $t$ are given on an interval $[0, b]$, where $b>0$. Suppose that $q(t)$ attains its minimum precisely at $t=0$, with $q^{\prime}(0)=0$, and that $q^{\prime \prime}(0)>0$. For a real argument $x$, define

$$I(x)=\int_{0}^{b} p(t) e^{-x q(t)} d t$$

Explain how to obtain the leading asymptotic behaviour of $I(x)$ as $x \rightarrow+\infty$ (Laplace's method).

The modified Bessel function $I_{\nu}(x)$ is defined for $x>0$ by:

$$I_{\nu}(x)=\frac{1}{\pi} \int_{0}^{\pi} e^{x \cos \theta} \cos (\nu \theta) d \theta-\frac{\sin (\nu \pi)}{\pi} \int_{0}^{\infty} e^{-x(\cosh t)-\nu t} d t .$$

Show that

$$I_{\nu}(x) \sim \frac{e^{x}}{\sqrt{2 \pi x}}$$

as $x \rightarrow \infty$ with $\nu$ fixed.