how to evaluate a sum of powers of central binomial coefficients?

Question

I have a series I was wondering how to evaluate. It has a fun result and I am wondering how one deals with sums of ,or alternating sums of, central binomial coefficients if they're cubed.

i.e. $\displaystyle \sum_{k=0}^{\infty}(-1)^{k}\left[\binom{2k}{k}x^{k}\right]^{3}$

Say, $x=1/4$. Then, we would have:

$\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^{k}[(2k)!]^{3}}{4^{3k}(k!)^{6}}$

According to Mathematica, this evaluates to $\displaystyle \frac{\pi}{\sqrt{2}\Gamma^{2}(5/8)\Gamma^{2}(7/8)}$

But, how can we evaluate this?.

I am familiar with several identities related to these, but not when they are to some power.

Perhaps Beta/Gamma can be implemented, but I am not sure how.

Thanks for any insight or advice.

EDIT: Apparently, these are related to Elliptic Integrals of the First Kind, and may be more difficult to evaluate than I had anticipated.

Answer 1

At the very least, you should easily recognize that you have a hypergeometric series, since

$$\binom{2k}{k}=\frac{\left(\tfrac12\right)_k4^k}{k!}$$

You are then asking about the hypergeometric function

$${}_3 F_2 \left({{\tfrac12,\tfrac12,\tfrac12}\atop{1,1}}\mid-64x^3\right)$$

As it turns out, this particular case is known to have an expression in terms of the complete elliptic integral of the first kind $K(m)$:

$${}_3 F_2 \left({{\tfrac12,\tfrac12,\tfrac12}\atop{1,1}}\mid z\right)=\frac4{\pi^2}K\left(\frac{1-\sqrt{1-z}}{2}\right)^2$$

This is a bit tedious to prove, so I'll omit it for now.

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To explain the expression in terms of gamma functions, it is known that for special values of the parameter $m$, called singular values, that the complete elliptic integral of the first kind is expressible in terms of gamma functions.

For $x=\tfrac14$, we have the expression

$$\frac4{\pi^2}K\left(\frac{1-\sqrt 2}{2}\right)^2$$

The argument of the elliptic integral is not within $(0,1)$, so the imaginary modulus transformation is necessary:

$$\frac4{\pi^2}K\left(\frac{1-\sqrt 2}{2}\right)^2=\frac4{\pi^2\left(1-\frac{1-\sqrt 2}{2}\right)}K\left(\frac{\tfrac{1-\sqrt 2}{2}}{\tfrac{1-\sqrt 2}{2}-1}\right)^2=\frac{8(\sqrt 2-1)}{\pi^2}K\left(3-2\sqrt 2\right)^2$$

Using formula 4 here (corresponding to the singular value $k_2$), we have

$$\frac{8(\sqrt 2-1)}{\pi^2}K\left(3-2\sqrt 2\right)^2=\frac{8(\sqrt 2-1)}{\pi^2}\frac{(\sqrt 2+1)\Gamma\left(\tfrac18\right)^2\Gamma\left(\tfrac38\right)^2}{2^{13/2}\pi}=\frac{\Gamma\left(\tfrac18\right)^2\Gamma\left(\tfrac38\right)^2}{2^{7/2}\pi^3}$$