NOTE: The 'correct' solution in this post is actually incorrect; see Cye Waldman's correct solution.
I am asked to solve the following integral for $E_n$, which appears in the study of the quartic anharmonic oscillator:
$$\int_{-E_n^{\frac{1}{4}}}^{E_n^{\frac{1}{4}}}\frac{dx}{\sqrt{E_n-x^4}} = 2\int_0^{E_n^{\frac{1}{4}}}\frac{dx}{\sqrt{E_n-x^4}} = \left(n+\frac{1}{2}\right)$$
The equality follows because the integrand is even. I now use the substitution $u=E_n-x^4$ to obtain:
$$dx=-\frac{du}{4(E_n-u)^\frac{3}{4}}$$
And the integral now becomes:
$$\frac{1}{2} E_n^{-\frac{3}{4}} \int_{0}^{E_n}u^{-\frac{1}{2}}(1-\frac{u}{E_n})^{-\frac{3}{4}}du$$
Making another substitution $v=\frac{u}{E_n}$, we obtain:
$$\frac 1 2 E_n^{-\frac{1}{4}} \int_0^1 v^{-\frac{1}{2}}(1-v)^{-\frac{3}{4}} \, dv$$
This is now a Beta function, and thus we obtain:
$$\frac{1}{2} E_n^{-\frac{1}{4}} \int_0^1 v^{-\frac{1}{2}}(1-v)^{-\frac{3}{4}}\,dv = \frac{1}{2} E_n^{-\frac{1}{4}} \ B(\frac{1}{2},\frac{1}{4}) = (n+\frac{1}{2})$$
Thus, rearranging for $E_n$, and writing the Beta function using Gamma functions, we obtain:
$$\implies E_n = \left(\frac{\Gamma(1/2) \Gamma(1/4)}{2 \Gamma(3/4) (n+\frac{1}{2})} \right)^4$$
However, the solution quoted by the question-setter, is:
$$E_n = \left(\frac{3\Gamma(3/4)^2}{\sqrt{2 \pi}}(n+\frac{1}{2}) \right)^{4/3}$$
This is drastically different. The identity:
$$\Gamma(1/4) \Gamma(3/4) = \sqrt{2} \pi$$
Is given, but this obviously would not resolve the discrepancy. I'm reasonably certain the error lies in one of my variable changes, but I cannot find where.
I am further inclined to think my answer is incorrect, as if the energy $E_n$ were indeed:
$$\implies E_n = \left(\frac{\Gamma(1/2) \Gamma(1/4)}{2 \Gamma(3/4) (n+\frac{1}{2})} \right)^4$$
Then $\displaystyle \lim_{n \to \infty} E_n = 0$, and the potential is bounded with a finite binding energy equal to the $n=0$ energy: $E_0 = \left(\frac{\Gamma(1/2) \Gamma(1/4)}{2 \Gamma(3/4) (\frac{1}{2})} \right)^4$. In other words, a particle could 'escape' the $x^4$ potential well, which does not make intuitive sense as it increases to infinity on both sides.
Best Answer
I disagree with correct solution in the OP and will present my own analysis for your scrutiny. First, let's reduce the clutter and set $a=E_n$. We seek to find $a$ such that
$$2\int_0^{a^{1/4}}\sqrt{a-x^4}dx=\left(n+\frac{1}{2}\right)$$
Now let $x^4=at$ or $x=(at)^{1/4}$, then
$$ dx=\frac{a^{1/4}}{4}t^{-3/4}dt\\ \sqrt{a-x^4}=\sqrt{a}\sqrt{1-t}\\ x=a^{1/4}\to t=1 $$
Substituting and rearranging we get
$$\frac{a^{3/4}}{2}\int_0^1 t^{-3/4}(1-t)^{1/2}dt=\left(n+\frac{1}{2}\right)$$
Introducing the complete beta function,
$$B(\nu,\mu)=\int_0^1 t^{\nu-1}(1-t)^{\mu-1}dt=\frac{\Gamma(\nu)\Gamma(\mu)}{\Gamma(\nu+\mu)}$$
Clearly, $\nu=1/4$ and $\mu=3/2$ and we can then show that
$$a=\left[ \frac{2\left(n+\frac{1}{2}\right)\Gamma(7/4)}{\Gamma(1/4)\Gamma(3/2)}\right]^{4/3}$$
At the OP's suggestion, we can substitute
$$ \Gamma(7/4)=(3/4)\Gamma(3/4)\\ \Gamma(3/2)=\sqrt{\pi}/2\\ \Gamma(1/4)\Gamma(3/4)=\pi\sqrt{2} $$
and demonstrate that
$$E_n=\left[ \frac{2\Gamma(3/4)^2\left(n+\frac{1}{2}\right)}{\pi\sqrt{2\pi}}\right]^{4/3}$$
So, it would appear that the original correct solution was missing a factor of $\pi$ in the denominator.
The solution I present here has been validated numerically for $n\in\mathbb{R^+}$, i.e., not just $n\in\mathbb{Z}$.