2

thanks in advance!

Calculate the leading-order asymptotic behaviour of the integral $$I(x) = \int_{0}^{2\pi} (1+t^2) e^{x \cos t} dt \mbox{ as } x \mbox { tends to infinity}$$

So far I know there are two maximas at $0 \mbox{ and }2\pi$ so it can be assumed to be the integral

$$\int_{0}^{\epsilon} (1+t^2) e^{x \cos t} dt + \int_{2\pi-\epsilon}^{2\pi} (1+t^2) e^{x \cos t} dt$$ then using taylor expansions this can be transformed to: $$\int_{0}^{\epsilon} (1+t^2) e^{x (1-\frac{t^2}{2})} dt + \int_{2\pi-\epsilon}^{2\pi} (1+t^2) e^{x (1-\frac{1}{2}(t-2\pi)^2)} dt$$ but where do I go from here? How do I find the leading order term?

Daniel
  • 417

1 Answers1

1

Basic outline: Also take the leading-order behavior of $1+t^2$ at the point of interest. So replace it with $1$ in the left integral and $1+(2\pi)^2$ in the right integral. Then replace $\epsilon$ with $\infty$ in both integrals.

Antonio Vargas
  • 25,541