# 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?

• Yes. Of course there is. And Cleo will post it in a few days, without any hint or comment whatsoever. Then you can write another post, asking how on earth she deduced either of the two results. :-) BTW, this reminds me of something our Analysis II teacher told us in college: he and his colleagues were very curious and scared of certain dreadful theorems, so they got their hands on a book. A manual, from their older friends. They looked it up. It had three points. The demonstrations were as follows: $1$. Obvious. $2$. Immediate. $3$. It follows from the previous two. :-) Nov 20, 2013 at 19:09
• The links are not working. And Lucian's comment suggests the answer is already given ?
– mick
Dec 13, 2013 at 23:08
• Using the definition of the G-function, we arrive at the following integral representation: $$A=2\sqrt2\pi\int^{\infty}_{-\infty}2^{6it}\Gamma(1/4+it)^2\sec^2(\pi(1/4+it))dt.$$ Apr 2, 2014 at 16:33

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 $$A(a)=2\pi^2\left(K_0(2a)-\left.\frac{d}{d\nu}L_\nu(2a)\right|_{\nu=0}\right).$$