Asymptotic Expansion of eigenvalues

Question

I would like to answer the following question

Show that the large eigenvalues of the system $$ \frac{d^2 w}{d x^2}=\left(x^4+x^2-\lambda^2\right) w, \quad w(-\infty)=w(\infty)=0, $$ are given by $$ \lambda \sim 2^{\frac{1}{3}} \pi\left(\Gamma\left(\frac{1}{4}\right)\right)^{-\frac{4}{3}}\left(3 n+\frac{3}{2}\right)^{\frac{2}{3}}+O\left(n^{\frac{1}{3}}\right), $$ where $n$ is a positive integer.

I understand that this is an eigenvalue problem with two turning points, namely $$ x_\pm = \pm \sqrt{\frac{1 + \sqrt{1+4 \lambda ^2}}{2}}. $$ The only way I know how to tackle this is by a WKB estimate which, in the limit $n \to \infty$ should give $$ (n+\frac{1}{2})\pi = \int_{x_-}^{x^+} \sqrt{\lambda^2-x^4-x^2} d x. $$ Now I am unsure how to proceed. I would think that here I would need to use some substitution and then seek some closed form solution from which I can invert for $\lambda$ in terms of $n$ and then do a regular expansion. However, I am not sure such closed form exists as I seem to be struggling with Wolfram saying that the integral is, by it self, $$ \frac{1}{6} \left(2x\sqrt{1-x^2-x^4}-i\sqrt{2(-1+\sqrt{5})}\, \text{EllipticE}\left[i\text{ArcSinh}\left(\sqrt{\frac{2}{1+\sqrt{5}}}x\right),-\frac{3}{2}-\frac{\sqrt{5}}{2}\right]+i(-5+\sqrt{5})\sqrt{\frac{2}{-1+\sqrt{5}}}\, \text{EllipticF}\left[i\text{ArcSinh}\left(\sqrt{\frac{2}{1+\sqrt{5}}}x\right),-\frac{3}{2}-\frac{\sqrt{5}}{2}\right]\right) $$ which does not look very pleasant.

Question: How does one proceed with this?

Answer 1

Hints:

1. Notice that $x_{\pm}\approx \pm\lambda^{1/2}$ if $\lambda\gg 1$.
2. Make the substitution $x=\lambda^{1/2}u$ in the WKB estimate and retain only the leading order term in the expansion in powers of $\lambda$: \begin{align} \left(n+\frac{1}{2}\right)\pi&\sim \int_{-1}^1\sqrt{\lambda^2(1-u^4)-\lambda u^2}\,\lambda^{1/2}du \\ &\sim\lambda^{3/2}\int_{-1}^1\sqrt{1-u^4}\,du +O(\lambda^{1/2})\\ &\sim\lambda^{3/2}\,\frac{\sqrt{\pi}\,\Gamma(5/4)}{\Gamma(7/4)}. \tag{1} \end{align} (Integral evaluated by WolframAlpha.)
3. Using properties of the gamma function, show that $$ \frac{\sqrt{\pi}\,\Gamma(5/4)}{\Gamma(7/4)} =\frac{[\Gamma(1/4)]^2}{3\sqrt{2\pi}}. \tag{2} $$
4. Substitute $(2)$ in $(1)$ and solve for $\lambda$.