Consider the following Bessel's equation $$\ddot{x}(t) + \frac{3}{t}\dot{x}(t) + x(t) = 0$$ with initial condition $x(0)=1$ and $\dot{x}(0)=0$. From WolframAlpha, we know the solution is given by $$x(t) = \frac{2J_1(t)}{t}$$ where $J_1(t)$ is the Bessel's function of the first kind with degree $1$. It is easy to see that $x(t)\to 0$ as $t\to\infty$. My question is that can we prove the trajectory of the solution $x(t)$ has finite length, i.e., $$\int_0^\infty |\dot{x}(t)| dt <\infty$$
I find that the trajectory is oscillating, so it seems that the trajectory length is pretty long, but since we have extra $1/t$ factor, maybe it makes the solution converge fast enough so that the length is indeed finite. Any ideas to prove or disprove would be helpful.
My trial:
\begin{align} \int_0^\infty |\dot{x}(t)| dt &= \int_0^\infty \left|\frac{2J_1'(t)}{t}-\frac{2J_1(t)}{t^2}\right|dt \\ &= \int_0^\infty \left|\frac{J_0(t)-J_2(t)}{t}-\frac{2J_1(t)}{t^2}\right|dt \end{align} Thus, a sufficient condition for the above quantity to be finite is $$\frac{J_0(t)}{t},\; \frac{J_2(t)}{t} \text{ and } \frac{J_1(t)}{t^2} \text{ are integrable on } \mathbb{R}_+.$$ From WolframAlpha I know $|J_1(t)|\leq 1$ for all $t\in\mathbb{R}_+$, so the last term above is obviously integrable. It suffices to check the first two are integrable. I guess $J_0(t)$ and $J_2(t)$ will have similar behavior at infinity, but I don't know how that implies integrability of $J_0(t)/t$ and $J_2(t)/t$.