I'm working through the exercises of Bender and Orszag's famous book, but I got stuck in 6.25 (a), in which it is asked to prove that
$$J_\nu (x) \sim (x/2)^\nu / \Gamma(\nu+1) \; \text{as} \; \nu \rightarrow \infty,$$
by using the following integral representation
$$J_\nu(x)=\frac{(x/2)^\nu}{\sqrt{\pi}\Gamma(\nu+1/2)} \int^\pi_0 \cos(x\cos\theta) \sin^{2\nu}\theta \, d\theta,$$
which is valid for $\nu > -1/2.$ ($J_\nu(x)$ is the $\nu$th-order Bessel function of the first kind.)
As the exercise belongs to section 6.4, which deals with Laplace's method and Watson's lemma, I thought I first had to perform a change of variables in order to get an integral of the form
$$I(x)=\int^b_a f(t)e^{x\phi(t)} \, dt.$$
So, I took $t=\cos\theta$ and obtained
$$\frac{(x/2)^\nu}{\sqrt{\pi}\Gamma(\nu+1/2)} \int^{1}_{-1} (1-t^2)^{p-\frac{1}{2}} e^{ixt} \, dt.$$
However, I cannot apply either Laplace's method or Watson's lemma, because the function $\phi$ I got is complex: $\phi(t)=it$.
What am I missing?
To begin, rewrite the integral as
$$ \int_0^\pi \cos(x\cos \theta) \exp\Bigl[2\nu \log \sin \theta\Bigr]\,d\theta. $$
The quantity $\log \sin \theta$ has a maximum at $\theta = \pi/2$, and near there
$$ \log\sin\theta = -\frac{1}{2} \left(\theta - \frac{\pi}{2}\right)^2 + \cdots. $$
Further
$$ \cos(x\cos\theta) = 1 + \cdots $$
there, so by the Laplace method we have
$$ \int_0^\pi \cos(x\cos \theta) \exp\Bigl[2\nu \log \sin \theta\Bigr]\,d\theta \sim \int_{-\infty}^{\infty} 1\cdot \exp\left[-2\nu \cdot \frac{1}{2} \left(\theta - \frac{\pi}{2}\right)^2\right]\,d\theta $$
for large $\nu$. Now simplify.