If have a question about section 3.1 in the paper Kramers' law: Validity, derivations and generalisations by Nils Berglund. (See http://arxiv.org/abs/1106.5799 page 7 - 9)
On page 8 it says, that the expected first-hitting time $w_A(x)=E^x\{\tau_A\}$ of $A$ can be achieved for a double well potential $V$ by solving:
$$ L w_A(x)=-1 \qquad x \in A^c \\
w_A(x)=0 \qquad x\in A $$
with L being the infinitesimal generator of the diffusion of the form $dx_t=-\nabla V(x_t)dt+\sqrt{2 \epsilon}dW_t$.
On Page 9 is the one-dimensional case considered with $(Lu)(x)=\epsilon u''(x)-V'(x)u'(x)$ and $A=(-\infty,a)$, $B=(b,\infty)$ for some $a<b$ and $x \in (a,b)$. And the solution is given by
$$ w_A(x)=\frac{1}{\epsilon} \int^x_a \int^{\infty}_z e^{[V(z)-V(y)]/\epsilon} dy \, dz $$
I don't know how to solve the problem to get the solution $w_A(x)$ for the one-dimensional case.
I hope someone can help me. Thanks in advance!
You just use an integrating factor to solve for $u'$ and then integrate.