# Limit $\lim\limits_{x\to\infty}\int\limits_0^\pi\cos\left(x\sin\theta \right)d\theta=0$

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.

• Would you allow to do a sub in the integral? That is, subbing away the sine, integrate wrt theta and then consider the limit? Commented Dec 2, 2019 at 0:30
• Maybe this helps? Commented Dec 2, 2019 at 0:33
• @imranfat yes how to do it? Commented Dec 2, 2019 at 0:45
• math.stackexchange.com/questions/2093915/…
– user140541
Commented Dec 2, 2019 at 1:09
• the solution from the ode is: $f(x) = c_1 J(x)+c_2Y(x)$ and both of the Bessel functions go to zero as $x\to 0,$ so I guess that is a proof. Commented Dec 2, 2019 at 1:16

$$f(x) = 2 \int_0^{\pi/2} \cos(x\sin\theta)\;d\theta$$ 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: $$\lim_{x \to \infty} \int_0^1\frac{\cos(xt)}{\sqrt{1-t^2}}\;dt = 0 .$$
• i think when you split up the interval, $\sin(\theta+\frac \pi 2) = \cos(\theta)$ Commented Dec 2, 2019 at 1:27
• @dezdichado: $\sin(\pi-\theta) = \sin(\theta)$. Commented Dec 2, 2019 at 1:28
• you are reducing the interval from $[0,\pi]$ to $[0,\pi/2]$ by that substitution? Commented Dec 2, 2019 at 1:29
• I use that substitution to get $\int_{\pi/2}^\pi \cos(x\sin\theta)d\theta = \int_{0}^{\pi/2} \cos(x\sin\theta)d\theta$ Commented Dec 2, 2019 at 1:31