I want to prove the following limit: $$\lim_{x\longrightarrow \infty} \frac{\int_0 ^{2\pi}e^{\sqrt{u^2+(v+x)^2}\cos\theta }d\theta}{\int_0 ^{2\pi}e^{x\cos\theta }d\theta}=e^v$$
where $u,v\in\mathbb{R}$. I got the result using Wolfram Mathematica but I have no idea how to prove this.
thanks for any help.
Computing the integrals, we obtain that the ratio is effectively equal to $$ \frac{I_0(\sqrt{u^2+(v+x)^2})}{I_0(x)}\ , $$ where $I_0$ is a Bessel function. Using the known asymptotics $$ I_0(y)\sim \frac{e^y}{\sqrt{2\pi y}}\ , $$ you are effectively after the limit $$ \lim_{x\to\infty} \frac{e^{\sqrt{u^2+(v+x)^2}}}{e^x}\frac{\sqrt{x}}{(u^2+(v+x)^2)^{1/4}}\ . $$ Given that $$ \lim_{x\to\infty} \frac{\sqrt{x}}{(u^2+(v+x)^2)^{1/4}}=1\ , $$ as both numerator and denominator are $\sim x^{1/2}$, then the limit to compute is $$ \lim_{x\to\infty} \frac{e^{\sqrt{u^2+(v+x)^2}}}{e^x}\ . $$ Since $\sqrt{u^2+(v+x)^2}\sim x+v$ for large $x$, it follows that $$ \frac{e^{\sqrt{u^2+(v+x)^2}}}{e^x}\sim \frac{e^{x+v}}{e^x}\to e^v\ . $$