In WKB approximation we expand the action in powers of $\hbar$ $$S(x)=S_0(x)+\dfrac{\hbar}{i}S_1(x)+\left(\dfrac{\hbar}{i}\right)^2S_2(x)+...$$ In standard treatment only terms up to and including $S_1(x)$ are used. In such a case, the transmission coefficient $T$ of tunneling across a potential barrier is given by
$$T=\exp\left(-\dfrac{2}{\hbar}S\right)$$
where $S$ is given by $$S=\int_{x_1}^{x_2}\sqrt{2m[V(x)-E]}dx$$
$V(x)$ is the potential energy, $E$ is the energy, $x_1$ and $x_2$ are the turning points satisfying $V(x_1)=E=V(x_2)$, and $m$ is the mass.
I read in references (see here and here) that $S_2(x)$ is given by
$$S_2(x)=\int \left(\dfrac{p''(x)}{4[p(x)]^2}-\dfrac{3}{8}\dfrac{[p'(x)]^2}{[p(x)]^3}\right)dx$$
where $p(x)=\sqrt{2m[E-V(x)]}$. Another equivalent formulation is $$S_2(x)=\int \left(\dfrac{V''(x)}{8\sqrt{2m}[E-V(x)]^{3/2}}+\dfrac{5 (V'(x))^2}{32\sqrt{2m}[E-V(x)]^{5/2}}\right)dx$$
My question is how $S$ (or more generally, $T$) will change if we include the second-order term $S_2(x)$ in WKB approximation?
My guess was that $S$ will change as
$$S=\int_{x_1}^{x_2}\sqrt{2m[V(x)-E]}+\hbar^2\int_{x_1}^{x_2}\left( \dfrac{V''(x)}{8\sqrt{2m}[V(x)-E]^{3/2}}+\dfrac{5 (V'(x))^2}{32\sqrt{2m}[V(x)-E)]^{5/2}}\right)dx$$
However, this does not seem right because the second and third term diverge/blow up at $x_1$ and $x_2$ where $V(x)-E$ becomes zero; hence the integration of these two integrands will be significantly bigger than the integration of the first one (I confirmed this numerically by trying out various potentials $V(x)$. For instance, I found that when $\int_{x_1}^{x_2}\sqrt{2m[V(x)-E]}dx \approx40$, the contribution from the other two integrals was $10^{23}$). This is at odds with my expectation in which when we incorporate $S_2(x)$, its contribution should be smaller than that of $S_0(x)$ ($\int_{x_1}^{x_2}\sqrt{2m[V(x)-E]}dx$).
So can anyone help me with that?
