How to prove $H\left(\frac{1}{4}\right)=\frac{e^{\frac{C}{4\pi}-\frac{3}{32}}\cdot A^{\frac{9}{8}}}{\sqrt 2}$

Question

How to prove: $$ H\left(\frac{1}{4}\right)=\frac{e^{\frac{C}{4\pi}-\frac{3}{32}}\cdot A^{\frac{9}{8}}}{\sqrt 2} $$ Where $C$ is Catalan's number, $A$ is Glaisher-Kinkelin's constant and $H(x)$ is the hyperfactorial given by: $$ H(n)=\prod_{k=1}^{n} k^k $$ $$ H(x-1)=\lim_{n\to\infty} \frac{e^{\frac{1}{2}x(x+1)}\cdot n^{xn+\frac{1}{2}x(x+1)}\cdot H(n)}{x^x\cdot (1+x)^{1+x}\cdots(n+x)^{n+x}} $$

Answer 1

We will use $(3)$ from here.

$$H(z-1)G(z)=e^{(z-1)\ln\Gamma(z)},$$

where $G$ is the Barnes G-function and $\Gamma$ is the Gamma function.

In our special case

$$H\left(\frac 14\right)=\frac{e^{\frac 14 \ln \Gamma\left(5/4\right)}}{G\left(\frac 54\right)}.$$

The easy part of the problem is to see that

$$e^{\frac 14 \ln \Gamma\left(5/4\right)}=\frac{\sqrt 2}{2} \Gamma\left(\frac 14\right)^{1/4}.$$

From J. Choi and H. M. Srivastava Certain Classes of Series Involving the Zeta Function from page $92$ to $94$ we could derive that

$$G\left(\frac 54\right) = e^{3/32\,-\,G/(4\pi)}\,A^{-9/8}\,\Gamma\left(\frac 14\right)^{1/4},$$

here $G$ denotes Catalan's constant, and $A$ is the Glaisher–Kinkelin constant.

From here it comes that

$$H\left(\frac{1}{4}\right)=\frac{e^{\frac{G}{4\pi}-\frac{3}{32}}\cdot A^{\frac{9}{8}}}{\sqrt 2}.$$