asymptotical behavior of integral

Question

I'm interest in the asymptotical of $$\int_{-\pi}^{\pi}\exp\Big((\cos z+i\alpha\sin z-1)t\Big)dz\hspace{3mm}\text{as}\hspace{2mm}t\to\infty$$ for $-1<\alpha<1$. Numberical result suggest that for $\alpha =0$ the integral behave like $\frac{c}{\sqrt{t}}$ and for $\alpha\neq 0$ it behave like $\exp(ct)$ for some $c<0$. I don't think Laplace method work for this problem. Thank you to suggest other method.

Answer 1

When $\alpha \neq 0$ you'll need to deform the contour over the new saddle point located at $z = i\operatorname{arctanh} \alpha$. At this saddle point we have

$$ \cos z + i\alpha \sin z = \sqrt{1-\alpha^2} - \frac{1}{2}\sqrt{1-\alpha^2} (z-i\operatorname{arctanh} \alpha)^2 + O(z-i\operatorname{arctanh} \alpha)^3, $$

so the Laplace method yields

$$ \begin{align} \int_{-\pi}^{\pi} \exp\Bigl[(\cos z + i \alpha \sin z)t\Bigr]\,dz &\sim \exp\left(t\sqrt{1-\alpha^2}\right) \int_{-\infty}^{\infty} \exp\left[-\frac{tz^2}{2}\sqrt{1-\alpha^2}\right]\,dz \\ &= \left(1-\alpha^2\right)^{-1/4} \sqrt{\frac{2\pi}{t}} \exp\left(t\sqrt{1-\alpha^2}\right). \end{align} $$

Thus

$$ \int_{-\pi}^{\pi} \exp\Bigl[(\cos z + i \alpha \sin z - 1)t\Bigr]\,dz \sim \left(1-\alpha^2\right)^{-1/4} \sqrt{\frac{2\pi}{t}} \exp\left[\left(\sqrt{1-\alpha^2}-1\right)t\right]. $$