A definite integral is defined as $$I(v,\theta)=\int_0^{\pi} e^{v[\cos(\theta-\phi)-1]}\sqrt{\dfrac{v \sin\phi}{\sin\theta}}d\phi$$
My question is how to show that $$\lim_{v\to \infty} I(v, \theta)=\sqrt{2\pi}$$ for any given $\theta \in (0, \pi)$.
My understanding is like this. Since $v$ goes to infinity, the main contribution of the integral comes from those $\phi$ values which are close to $\theta$ (because of the exponential term). I tried the Taylor expansion of $\phi$ around $\theta$, for example, \begin{align*} \cos (\theta -\phi) &\approx 1-\dfrac{(\theta-\phi)^2}{2!} + \dfrac{(\theta-\phi)^4}{4!} + \cdots \\ \sin \phi & \approx \sin \theta + \cos\theta(\phi-\theta) - \dfrac{\sin \theta}{2}(\phi-\theta)^2 + \cdots \end{align*} but still can't go far. I don't know if I am on the right track. Please help.