TITLE: Limit $\lim\limits_{x\to\infty}\int\limits_0^\pi\cos\left(x\sin\theta \right)d\theta=0$
QUESTION [6 upvotes]: Consider the function $f$ defined on $\mathbb{R}$ by
$$f(x)=\int_0^\pi\cos\left(x\sin\theta  \right)d\theta.$$
I showed that this function satisfies the following differential equation:
$$xf''(x)+f'(x)+xf(x)=0$$
this implies that
$$f''(x)+\frac{f'(x)}{x} +f(x)=0$$
since $f'$ is bounded then $$\lim_{x\to\infty}f''(x)+f(x)=0$$
how to continue to prove that $$\lim_{x\to\infty}f(x)=0.$$
I am also interested if there's another method to prove it without the differential equation.

REPLY [6 votes]: $$
f(x) = 2 \int_0^{\pi/2} \cos(x\sin\theta)\;d\theta
$$
change variables, $t = \sin\theta$
$$
f(x) = 2 \int_0^1\frac{\cos(xt)}{\sqrt{1-t^2}}\;dt
$$
But $\frac{1}{\sqrt{1-t^2}}$ is integrable on $[0,1]$, so we may conclude from the Riemann-Lebesgue lemma that:
$$
\lim_{x \to \infty} \int_0^1\frac{\cos(xt)}{\sqrt{1-t^2}}\;dt = 0 .
$$