I am trying to study the function $$ V(x)=\sum_{n=0}^{\infty}\frac{(4n)!}{n!^4}\frac{x^{2n}}{16^{2n}} $$
Using the Beta function, one can prove that $$ V(x)=\frac{4}{\pi^2}\int_0^1 \frac{K(xt^2)}{\sqrt{1-t^2}}\, dt=\frac{2}{\pi^2}\int_0^1 \frac{K(xt)}{\sqrt{t(1-t)}}\, dt $$
My solution to $V(1)$ is as the following:
substitute $t=\frac{2\sqrt u}{1+u}$ to the last integral, $ dt= \frac{1-u}{\sqrt{u}(1+u)^2}\, du$, this gives \begin{equation}\label{pseudo landen int. rep.} V(x)=\frac{\sqrt 2}{\pi^2}\int_0^1 \left(\frac{1+\sqrt{u}}{1+u}\right)K\left(\frac{2x \sqrt{u}}{1+u}\right)\frac{du}{u^{3/4}} \end{equation} Setting $x=1$ gives $$V(1)=\frac{\sqrt 2}{\pi^2}\int_0^1 \left(\frac{1+\sqrt{u}}{1+u}\right)K\left(\frac{2 \sqrt{u}}{1+u}\right)\frac{du}{u^{3/4}}$$
The Landen transformation of Elliptic integral gives $K(k)=\frac{1}{1+k}K(\frac{2\sqrt{k}}{1+k})$,
$$V(1)=\frac{\sqrt{2}}{\pi^2}\int_0^1 \frac{1+\sqrt u}{u^{3/4}}K(u)\, du$$ substitute $u=v^4$ gives
$$=\frac{\sqrt{2}}{\pi^2}\int_0^1 \frac{1+v^2}{v^{3}}K(v^4)\,4v^3 \, dv=\frac{4\sqrt 2}{\pi^2} {\underbrace{\int_0^1 (1+v^2) K(v^4)\, dv}_{I}}$$
$$I=\int_0^1 (1+k^2) K(k^4)\, dk$$
$$=\int_0^1 \int_0^1\frac{(1+k^2)}{\sqrt{(1-t^2k^8)(1-t^2)}}\, dt\, dk=\int_0^1 \int_0^1\frac{(1+k^2)}{\sqrt{(1-t^2k^8)(1-t^2)}}\, dk\, dt$$
substitute $kt^{1/4}= u$ , $t^{1/4} \, dk= du$ $$=\int_0^1 \int_0^{t^{1/4}}\frac{(1+(ut^{-1/4})^2)}{\sqrt{(1-u^8)(1-t^2)}}\, \frac{du}{t^{1/4}}\, dt$$ substitute $t=x^4$ , $dt= 4x^3\, dx$ $$=\int_0^1 \int_0^{x}\frac{(1+u^2x^{-2})}{\sqrt{(1-u^8)(1-x^8)}}\, \frac{du}{x}\, 4x^3\, dx=\int_0^1 \int_0^{x}\frac{4x^2+4u^2}{\sqrt{(1-u^8)(1-x^8)}}\, du \, dx$$
Switching integral signs gives $$\int_0^1 \int_u^1 \frac{4x^2+4u^2}{\sqrt{(1-u^8)(1-x^8)}}\, dx\, du$$
Notice that the integrand is symmetric, we add the two integrals together, giving
$$2I=\int_0^1 \int_0^{x}\frac{4x^2+4u^2}{\sqrt{(1-u^8)(1-x^8)}}\, du \, dx+\int_0^1 \int_u^1 \frac{4x^2+4u^2}{\sqrt{(1-u^8)(1-x^8)}}\, dx\, du$$ $$=\int_0^1 \int_0^{1}\frac{4x^2+4u^2}{\sqrt{(1-u^8)(1-x^8)}}\, du \, dx$$ $$=\int_0^1 \int_0^{1}\frac{4x^2}{\sqrt{(1-u^8)(1-x^8)}}\, du \, dx+\int_0^1 \int_0^{1}\frac{4u^2}{\sqrt{(1-u^8)(1-x^8)}}\, du \, dx$$ $$=2\int_0^1 \int_0^{1}\frac{4x^2}{\sqrt{(1-u^8)(1-x^8)}}\, du \, dx$$ We can finally separate the integrals $$\implies I=\int_0^1 \int_0^{1}\frac{4x^2}{\sqrt{(1-u^8)(1-x^8)}}\, du \, dx =4\left(\int_0^1 x^2(1-x^8)^{-1/2}\, dx\right)\left(\int_0^1 (1-u^8)^{-1/2}\, du\right)$$ substitute $x=t^{1/8}$ and $u=t^{1/8}$ $$=4\left(\int_0^1 t^{1/4}(1-t)^{-1/2}\cdot \frac{1}{8}t^{-7/8}\, dt\right)\left(\int_0^1 (1-t)^{-1/2}\cdot \frac{1}{8} t^{-7/8}\, dt\right)$$ $$=\frac{1}{16}\left(\int_0^1 t^{-5/8}(1-t)^{-1/2}\, dt\right)\left(\int_0^1 t^{-7/8} (1-t)^{-1/2}\, dt\right)$$ $$=\frac{1}{16}\cdot \frac{\Gamma(\frac{3}{8})\Gamma(\frac{1}{2})}{\Gamma(\frac{7}{8})}\cdot \frac{\Gamma(\frac{1}{8})\Gamma(\frac{1}{2})}{\Gamma(\frac{5}{8})}=\frac{\pi }{16}\cdot \frac{\Gamma(\frac{3}{8})\Gamma(\frac{1}{8})}{\Gamma(\frac{7}{8})\Gamma(\frac{5}{8})}$$ Using the Euler reflection formula, it gives $$=\frac{\pi }{16}\cdot \frac{1}{\Gamma^2(\frac{7}{8})\Gamma^2(\frac{5}{8})}\cdot\frac{\pi}{\sin(3\pi/8)}\frac{\pi}{\sin(\pi/8)}=\frac{\pi^3}{4\sqrt{2}\Gamma^2(\frac{7}{8})\Gamma^2(\frac{5}{8})}$$
$$\implies I=\frac{\pi^3}{4\sqrt{2}\Gamma^2(\frac{7}{8})\Gamma^2(\frac{5}{8})}$$
\begin{equation}\label{V1} V(1)=\frac{\pi}{\Gamma^2(\frac{7}{8})\Gamma^2(\frac{5}{8})} \end{equation}
This establishes a closed form of $V(1)$.
Now, the critical step of my derivation is the Landen transform part. It greatly simplifies the integrand. If $x$ is not $1$, I cannot do any Landen transform. Therefore, to gain more insight of $V(x)$, I would like a different approach to such computation. Maybe some that could find other values of $V(x)$.
Any help is apprciated.
