Simplification of $G_{2,4}^{4,2}\left(\frac18,\frac12\middle|\begin{array}{c}\frac12,\frac12\\0,0,\frac12,\frac12\\\end{array}\right)$ In this post Cleo gives a misterious result containing the following generalized Meijer G-function:
$$G_{2,4}^{4,2}\left(\frac18,\frac12\middle|\begin{array}{c}\frac12,\frac12\\0,0,\frac12,\frac12\\\end{array}\right)$$
Is it possible to represent it in terms of simpler (including hypergeometric) functions?
 A: Using the definition of the G-function, we get the following integral representation: 
$$A=2\sqrt2\int^{\infty}_{-\infty}2^{2it}\Gamma(\tfrac14-it)^2\Gamma(\tfrac12+2it)^2dt\approx37.1364$$
We write $F(z)=2^{2iz+\frac32}\Gamma(\tfrac14-iz)^2\Gamma(\tfrac12+2iz)^2$. $F(z)$ have poles at $z=(\tfrac{n}2+\tfrac14)i$ and $z=(-n-\tfrac14)i$ for all integer $n\geq0$.
We use a rectangular contour with vertices $-N,N,N+Ni,-N+Ni$, where $N$ is an integer. It is able to prove that the integral on the other three sides goes to zero as $N\to\infty$, so we have $$A=2\pi i\sum_{n=0}^{\infty}\operatorname{Res}\left(F,\left(\frac{n}2+\frac14\right)i\right).$$
We can show that 
$$\begin{align*}
2\pi i\operatorname{Res}\left(F,\left(\frac{n}2+\frac14\right)i \right)&=2\pi^2\frac{\psi\left(\frac{n}2+1\right)+3\log2}{2^{3n}\Gamma\left(\frac{n}2+1\right)^2}.
\end{align*}$$
So we have a series representation 
$$A=2\pi^2\sum^{\infty}_{n=0}\frac{\psi\left(\frac{n}2+1\right)+3\log2}{2^{3n}\Gamma\left(\frac{n}2+1\right)^2}.$$
It remains to convert this expression into an closed form involving hypergeometric functions.
Edit: There is a more general result, $$
A(a)=G_{2,4}^{4,2}\left(a,\frac12\middle|\begin{array}c\tfrac12,\tfrac12\\0,0,\tfrac12,\tfrac12\end{array}\right)=2\pi^2\sum^{\infty}_{n=0}\frac{a^n\left(\psi\left(\frac{n}2+1\right)-\log a\right)}{\Gamma\left(\frac{n}2+1\right)^2}.
$$
Using DLMF 10.31.1 and 10.25.2, we can prove that the sum of the even terms,
$$
A_{even}(a)=2\pi^2\sum^{\infty}_{n=0}\frac{a^{2n}\left(\psi(n+1)-\log a\right)}{\Gamma\left(n+1\right)^2}=2\pi^2K_0(2a).
$$
Curiously, this cancels the $K_0$ term in Cleo's answer.
Edit: The odd terms have a closed form related to derivative of the Struve functions, by using DLMF 11.2.2:
$$
A_{odd}(a)=2\pi^2\sum^{\infty}_{n=0}\frac{a^{2n+1}\left(\psi\left(n+\frac32\right)-\log a\right)}{\Gamma\left(n+\frac32\right)^2}=-2\pi^2\left.\frac{d}{d\nu}L_\nu(2a)\right|_{\nu=0}.
$$
Therefore, we have
$$A(a)=2\pi^2\left(K_0(2a)-\left.\frac{d}{d\nu}L_\nu(2a)\right|_{\nu=0}\right).$$
