Why is the WKB tunneling amplitude a non-perturbative result?

The tunneling amplitude obtained from WKB aprroximation is given by $$|T(E)|=\exp\Big\{-\frac{1}{\hbar}\int\limits_{x_1}^{x_2}dx[2(V(x)-E)]^{1/2}\Big\}[1+O(\hbar)]$$ where $x_1$ and $x_2$ are the classical turning points at energy $E$.

Why is this referred to as a non-perturbative result? Why can't this phenomenon reveal itself, as it is often said, at any order in perturbation theory?

Addendum: Given a non-solvable potential, I can solve for the approximate energy eigenfunctions, and compute the tunneling amplitude as one does for a step potential. Will it not reveal the tunneling phenomenon?

Consider the function $\exp\Big(-\frac{1}{\hbar}f(x)\Big)$. Every derivative of it can be written in the form, \begin{equation} \frac{d^n}{d\hbar^n}\exp\Big(-\frac{1}{\hbar}f(x)\Big)=\frac{p_{2n}\Big[\hbar,f(x)\Big]}{\hbar^{2n}}\exp\Big(-\frac{1}{\hbar}f(x)\Big) \end{equation} where $p_{2n}$ is some polynomial in $\hbar$. In the limit $\hbar\rightarrow 0$ the exponent goes to zero faster than any inverse power increases. Therefore for any $n$, \begin{equation} \frac{d^n}{d\hbar^n}\exp\Big(-\frac{1}{\hbar}f(x)\Big)\Bigg|_{\hbar=0}=0 \end{equation} Therefore Taylor series in $\hbar$ is zero at every order and such contribution won't be seen in any order of perturbation theory in $\hbar$ i.e. is non-perturbative for such expansion.
Now we considered the expansion in $\hbar$. However note that $\hbar$ is either some dimensional quantity or set to $1$. What you actually have to consider is some dimensionless combination of the model parameters that usually involves what we call a coupling constant.
• Perturbation theory relies on an expansion in terms of the coupling constant/strength of the perturbation whereas the WKB method relies on an expansion in powers of $\hbar$. I don't see any one-to-one correspondence between the two. Then how does one explain why WKB tunnelling amplitude is non-perturbative? @OON – SRS Dec 22 '17 at 13:05
• Usually we make perturbation expansion by orders of coupling. Here it's not pretty rigorous to write explicitly. But you can take "$\hbar$" as a example, say, the exponential term contains contributions from infinite number of $hbar$ power terms which it converges, but if you make perturbation expansion to the exponential term around $\hbar=0$ you cannot get higher order contributions to the amplitude, as the Taylor coefficients are all zero. If you make perturbative calculation directly to the barrier, then the transmission amplitude you get can't be exponential usually, right? – Kangle Aug 4 '17 at 15:37