Consider a stochastic differential equation:
$$dX_t = dW_t + \sin(X_t) dt, \, X_0 = x$$
where $W_t$ is a Wiener process. Define
$$\tau_1 = \inf \{ t : X_t \in 2 \pi \mathbb{Z} \} \\ \tau_2 = \inf \left \{ t : X_t \in 2 \pi \mathbb{Z} \setminus \left \{ X_{\tau_1} \right \} \right \}.$$
My goal is to show that $\tau_1$ and $\tau_2$ both have finite expectation; this will allow me to solve a larger problem about recurrence using very standard arguments.
Since the drift is $2 \pi$ periodic in space and homogeneous in time while the diffusion is homogeneous in both space and time, it is enough to consider $x \in (0,2 \pi)$ for the first part and $X_{\tau_1}=0$ for the second part. In the first case we will be done provided the problem
$$\frac{1}{2} u'' + \sin(x) u' = -1, \, u(0)=u(2\pi)=0$$
has a nonnegative solution. In the second case we will be done provided the problem
$$\frac{1}{2} u'' + \sin(x) u' = -1, \, u(-2\pi)=u(2\pi)=0$$
has a nonnegative solution. These equations may be solved in explicit form, but in terms of some non-elementary integrals, through reduction of order and integrating factors. For instance the solution to the first problem may be written as
$$u(x) = -2 w(x) + 2 \frac{w(2 \pi) v(x)}{v(2 \pi)}$$
where
$$v(x) = \int_0^x e^{2 \cos(y)} dy \\ w(x) = \int_0^x \int_0^y e^{2(\cos(y)-\cos(z))} dz dy. $$
If I can prove $u \geq 0$ then I will be done. I'm fairly sure that any proof that this $u \geq 0$ will work on the second one. Any suggestions?
Edit: a quick numerical check shows that $u \geq 0$ should indeed hold.
A simple approach (which avoids the task of checking the nonnegativity of the function in your question) is to compare the hitting times $\tau_1$ and $\tau_2$ with the hitting time $\theta$ of $\pm2\pi$ by the process $$dY=\mathrm dW-\mathrm{sgn}(Y)\,\mathrm dt,$$ starting from some $|y|\leqslant2\pi$. Since $|\sin|\leqslant1$, $\tau_1$ and $\tau_2$ are dominated stochastically by $\theta$, and $\theta$ starting from any $|y|\leqslant2\pi$ is stochastically dominated by $\theta$ starting from $0$. Finally, $s(y)=E_y(\theta)$ solves $\frac12s''-s'=-1$ with $s(\pm2\pi)=0$ and $s'(0)=0$ hence a standard computation shows that $$8s(x)=\mathrm e^{4\pi}-\mathrm e^{2|x|}+2(|x|-2\pi),$$ in particular, for every $x$, $$E_x(\tau_1)\leqslant E_x(\tau_2)\leqslant E_0(\theta)=\tfrac18(\mathrm e^{4\pi}-1-4\pi),$$ which is finite.