Variational Method 
A particle is moving in one dimension under a potential $V(x)$ such that, for large positive values of $x$, $V(x) \approx  kx ^\beta$, where $k>0$ and $\beta$ $\geq$ 1. If the wave function in this region has the form $\psi(x)\sim \exp(-x^\lambda)$, Then show that,
$$\lambda=\frac{\beta}{2}+1.$$


I am not sure how to approach this problem. I thought of using Variational Method, but ended up in a complicated Gaussian integral. Is there any other approach as to reach to this result.
 A: We can use one dimensional time-independent Schrödinger equation,
$$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\Psi(x)+V(x)\Psi(x) = E\Psi(x)$$
Given that
$\Psi(x)\approx e^{-{x^\lambda}}$ and $V(x)\approx kx^\beta$, so we get
$$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}e^{-{x^\lambda}}+kx^\beta e^{-{x^\lambda}} = Ee^{-{x^\lambda}}$$ which gives
$$-\frac{\hbar^2}{2m}\lambda[(\lambda-1)x^{\lambda-2}-\lambda x^{2(\lambda-1)}]+kx^{\beta}= E$$
Now we assume $\lambda>0$, then $O(x^{\lambda-2})<O(x^{2(\lambda-1)})$, so we may neglect the first term in the limit of large x. We can now write,
$$\frac{\hbar^2}{2m}\lambda^2 x^{2(\lambda-1)}+kx^{\beta}\approx E$$
Differentiating both sides wrt x we get
$$2\frac{\hbar^2}{2m}\lambda^2 (\lambda-1) x^{2(\lambda-1)-1}+k\beta x^{\beta-1}\approx 0$$
Now notice that the RHS of the equation is 0 while LHS is x dependent. So this equation can only be valid for all x if
$$2(\lambda-1)-1=\beta - 1$$ because if the two are not equal then both the x terms will be linearly independent and their linear combination cannot yield zero. Finally we get,
$$\lambda = \frac{\beta}{2}+1$$
