Eigenvalue problem for $−\psi''(x) − (ix)^ N \psi(x) = E\psi(x)$ in complex plane To find the eigenvalue $E$ in the complex plane of $x$ for one dimensional Schrodinger equation
$$
−\psi''(x) − (ix)^ N \psi(x) = E\psi(x).
$$
where $N$ can be any real number, the boundary condition  $\psi(x) \to 0$ as $|x| → ∞$ causes a great difficulty for numerical method, because there are infinite number of contour paths which go from zero to complex infinity. 
Bender et al on their paper only considered the contour paths (e.g. the black path on the following figure) which are entirely confined in two Stokes wedges symmetric with respect to the imaginary axis. Based on my understanding on what they say, each asymptotic series expansion for $ψ(x)$ is only valid within certain sector or wedge, therefore for any path which is outside of these two wedges, the corresponding asymptotic series is no longer asymptotic to the value of the function $ψ(x)$ . However, there exists a new, unique and valid asymptotic series for $ψ(x)$ outside the wedge. Yet, this new asymptotic series, say $f(x)$, does not satisfy the boundary condition  $f(x) \to 0$ as $|x| \to \infty$ .

My question:
I don't really understand why the contour path has to be entirely inside the Stokes sector. How about the red and yellow path shown on the figure above?
They are partly inside the Stokes wedges. The red path starts from the complex infinity which lies inside the right Stokes wedge and then approaches the origin zero from the left Stokes wedge, so that only the middle part of the red path lies outside the two Stokes sectors. The yellow path starts from the complex infinity which lies inside the right Stokes wedge and then approaches the origin zero inside the right Stokes wedge again.
Why do these red and yellow paths have to be excluded please? If we integrate along anyone of these two paths, does it yield finite eigenvalue and exponentially decayed eigenfunction or not? Why?
 A: It absolutely does not matter where the normalization contour starts and how it behaves in the beginning. It only matters how it behaves near infinity. The behavior of solution of your equation is
the following: for every Stokes sector, there is a one-dimensional set of solutions which tend to zero on any path "strictly" within this Stokes sector.
"Strictly" means that if the Stokes sector is $\alpha<\arg z<\beta$ then
the argument on the path $\gamma(t):0\leq t<\infty$ should satisfy 
$$\alpha<\liminf_{t\to\infty}\arg \gamma(t)<\limsup_{t\to\infty}\arg\gamma(t)<\beta-\epsilon,$$
for some $\epsilon>0$. What the path does for finite $t$ is irrelevant,
only $\arg\gamma(t)$ as $t\to\infty$.
So the boundary conditions are essentially determined by choosing two
non-adjacent Stokes sectors where solution tends to $0$, the contour is irrelevant and for computation you can choose whatever contour you wish. No other reasonable boundary conditions at $\infty$ can be imposed. 
So all three contours in your picture are fine provided that they stay in the green sector far away from the origin.
