WKB Quantization Condition - negative?

In deriving the quantization condition for a bound state in a potential with "no verticle walls" we start with the WKB connection formulas to find the wavefunction in the interior of the well ($x_1<x<x_2$), namely $$\psi(x) \cong \begin{cases}\displaystyle\frac{2C_1}{\sqrt{p(x)}}\sin\left[\frac{1}{\hbar}\int_{x_1}^x{dx\ p(x)}+\frac{\pi}{4}\right]&\quad\text{for}\quad x_1<x\\ \\\displaystyle \frac{2C_2}{\sqrt{p(x)}}\sin\left[\frac{1}{\hbar}\int_x^{x_2}{dx\ p(x)}+\frac{\pi}{4}\right]&\quad\text{for}\quad x<x_2 \end{cases}$$

Our task is to connect the two wave functions in the interior region together. Seeing that the coefficients are easily handled, it is then necessary that the arguments of the sine functions be equivalent. Denoting $$\theta_1(x) = -\frac{1}{\hbar}\int_{x_1}^x{dx\ p(x)}-\frac{\pi}{4}$$ $$\theta_2(x) = \frac{1}{\hbar}\int_{x}^{x_2}{dx\ p(x)}+\frac{\pi}{4}$$ we have $$\theta_1(x)+n\pi = \theta_2(x)\quad\text{for}\quad n = 0,1,2,3..$$ since the relative minus can be absorbed into either coefficient. Thus $$-\frac{1}{\hbar}\int_{x_1}^x{dx\ p(x)}-\frac{\pi}{4}+n\pi= \frac{1}{\hbar}\int_{x}^{x_2}{dx\ p(x)}+\frac{\pi}{4}$$ \begin{equation}\int_{x_1}^{x_2}{dx\ p(x)} = \left(n-\frac{1}{2}\right)\pi\hbar\quad\text{for}\quad n=0,1,2,3..\end{equation} This is our quantization condition for the allowed energies.

However, following the derivation in Griffiths or Sakurai, $n=0$ is not included. Why?

My ideas: Well this would imply $$\theta_1(x) = \theta_2(x)$$ \begin{equation}\int_{x_1}^{x_2}{dx\ p(x)} = -\frac{\pi}{2}\hbar\end{equation} \begin{equation}\int_{x_1}^{x_2}{dx\ \sqrt{2m(E-V(x))}} = -\frac{\pi}{2}\hbar\end{equation} I am trying to imagine why a negative value of the action integral might be prohibited. Am I correct in assuming that if we were to restrict all possible values of the energy to be real, it would that imply $$p(x)>0\$$ Can someone provide a proof as to why $$p(x)>0\rightarrow E\in\mathbb{R}.$$It would then follow the the integral of a strictly positive integrand must itself be positive, which means we cannot have $-\frac{\pi}{2}\hbar$.

1. Normally we assume that $-\infty< x_1 < x_2<\infty$ are the turning points in the 1D potential well. This means $$\forall x\in ]x_1,x_2[:~~ E ~>~ V(x).$$ Hence OP's third last equation $$0 ~<~ \int_{x_1}^{x_2}{dx\ \sqrt{2m(E-V(x))}} ~=~ -\frac{\pi}{2}\hbar~<~0$$ can never be fulfilled.