Conjectured closed form of $G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right)$ In my answer to this question, I come across the following case of the Meijer G-function:
$$F(b)=G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right), b>0$$
and based on my experiments, $F(b)$ have the following closed form:
$$F(b)
\stackrel?=\frac{\Gamma(b)}{b}\left(-\gamma-\psi(b)+\frac{2^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-1\right)+b{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-1\right)\right)$$
Is there any chance of proving this?
Edit: Using the definition of the Meijer G-Function, $F(b)$ have the integral representation:
$$F(b)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}dx.$$
Edit 2: I've found a further generalization:
$$F(b,z)=G^{2~2}_{3~3}\left(z\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}e^{(b/2+ix)\log z}dx\\
\stackrel?=\frac{\Gamma(b)z^b}{b}\left(-\log z-\gamma-\psi(b)+\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\right), b\not\in\{0,-1,-2,\dots\},z\neq0,-1.$$
Edit 3: Further simplified the ${_2F_1}$ part.
 A: Case $F(1,1)$ is this claim:
$$
2 \int_{-\infty}^\infty \frac{dx}{(4 x^2+1)\cosh(\pi x)} = 2\log 2 .
$$
Maybe start with a proof of that.  
added March 24 
We will attempt this $F(1,1)$ case using residues.  I changed variables slightly to make the problem
$$
\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)\cosh(\pi x/2)} = 2\log 2 .
$$
The function
$$
F(z) := \frac{1}{(z^2+1)\cosh(\pi z/2)}
$$
is analytic in the complex plane except for poles at $(2n+1)i$, $n \in \mathbb Z$.  Let $M$ be a large integer, and consider the contour integral of $F$ around the square with vertices
$$
-4 M ,\quad 4 M ,\quad 4M + 8 M i,\quad -4 M\pi + 8 M i .
$$
The integral along the bottom side is
$$
\int_{-4M}^{4M} \frac{dx}{(x^2+1)\cosh(\pi x/2)}
$$
so the limit as $M \to \infty$ is the integral to be computed.  We can do estimates for the other three sides to show their integrals go to zero as $M \to \infty$.  (Postponed...)  
The poles for $F(z)$ inside the square are at $z=(2n+1) i$, for $n=0,1,\dots, 4M-1$.  Now $z=i$ is a double pole, the residue there is
$$
\mathrm{Res}_{z=i} F(z) = \frac{-i}{2\pi} .
$$
For $n>0$, the residue is
$$
\mathrm{Res}_{z=(2n+1) i} F(z) = \frac{(-1)^n i}{2 n (n+1)\pi}
$$
Now as $M \to \infty$, the square encloses more and more of these poles, so in the limit the sum is the sum of all the residues in the upper half-plane:
$$
\frac{-i}{2\pi}+\sum_{n=1}^\infty \frac{(-1)^n i}{2n(n+1)\pi}
=\frac{-i\log 2}{\pi}
$$
so finally the integral around the contour (which converges to the integral we want) also converges to $2\pi i$ times this sum of residues,
$$
\int_{-\infty}^{+\infty} F(z)\,dz = 2 \log 2 .
$$
A: Using the method in the other answer, assuming it is enough to sum the residues in the upper half-plane as in that case, I get
$$
\frac{1}{2\pi}\,\int _{-\infty }^{\infty }\!{\frac {\Gamma  \left(b/2+ix
 \right) \Gamma  \left( b/2-ix \right) {{\rm e}^{ \left( b/2+ix
 \right) \ln  \left( z \right) }}}{{b}^{2}/4+{x}^{2}}}{dx}
={\frac {\Gamma  \left( b \right) \ln  \left( z \right) }{b}}-{
\frac {\Psi \left( b \right) \Gamma  \left( b \right) }{b}}-{\frac {
\gamma\,\Gamma  \left( b \right) }{b}}+{\frac {\Gamma  \left( b
 \right) }{{b}^{2}}}+
\frac{\Gamma  \left( b \right) b}{z(b+1)}
\;{\mbox{$_4$F$_3$}\left(1,1,b+1,b+1;\,2,2,b+2;\,-\frac{1}{z}\right)}
$$
Numerically, this seems to agree with the formula given in the question.
A: Inspired by GEdgar's answer for the $F(1,1)$ case, I give an answer (to my own question!) for the general case.
We start from the integral representation $$
F(b,z)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}z^{\frac{b}{2}+ix}dx.
$$
We consider the function $G(w)=\frac{\Gamma(\tfrac{b}{2}+iw)\Gamma(\tfrac{b}{2}-iw)}{\tfrac{b^2}{4}+w^2}z^{\frac{b}{2}+iw}dw$. $G(w)$ have double poles at $w=\pm\frac{b}{2}i$, and single poles at $w=\pm(\frac{b}{2}+n)i$ for all positive integer $n$. 
Now let $N$ be a large integer, and consider the integral path around the rectancle with vertices $-N,N,N-(N+\frac{b+1}{2})i,-N-(N+\frac{b+1}{2})i$. It is possible to prove that the integrals on the three sides go to zero, and therefore we have 
$$F(b,z)=-i(\text{sum of residues of $G(w)$ in the lower half-plane}).$$
It remains to calculate the residues at each pole.
In fact, we have $$
\begin{align*}
-i\operatorname{Res}_{w=-\frac{b}{2}i}G(w)&=\frac{z^b\Gamma(b)}{b}\left(\frac1b-\log z-\gamma-\psi(b)\right)\\
-i\operatorname{Res}_{w=-(\frac{b}{2}+n)i}G(w)&=\frac{(-1)^{n+1}z^{b+n}\Gamma(b+n)}{n~n!(b+n)}.
\end{align*}
$$
Comparing with the conjectured closed form, we only need to prove that $$
\frac1b-\sum_{n=1}^{\infty}\frac{(-z)^nb\Gamma(b+n)}{n~n!\Gamma(b)(b+n)}\stackrel?=\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right).$$
Now note that $$
\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\\
=b^{-1}{_2F_1}\left(\begin{array}cb,b\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\\
=\underbrace{\sum_{m=0}^{\infty}\frac{\Gamma(b+m)^2\Gamma(b+1)}{b\Gamma(b)^2\Gamma(b+m+1)m!}(-z)^m}_{n=m}-\underbrace{\sum_{m=0}^{\infty}\frac{b\Gamma(m+1)^2\Gamma(b+m+1)\Gamma(2)^2}{\Gamma(1)^2\Gamma(b+1)\Gamma(m+2)^2m!}(-z)^{m+1}}_{n=m+1}\\
=\sum_{n=0}^{\infty}\frac{\Gamma(b+n)}{\Gamma(b)(b+n)n!}(-z)^n-\sum_{n=1}^{\infty}\frac{\Gamma(b+n)}{n~n!\Gamma(b)}(-z)^{n}\\
=\frac1b+\sum_{n=1}^{\infty}\left(\frac{\Gamma(b+n)}{\Gamma(b)(b+n)n!}(-z)^n-\frac{\Gamma(b+n)}{n~n!\Gamma(b)}(-z)^{n}\right)\\
=\frac1b+\sum_{n=1}^{\infty}\frac{\Gamma(b+n)(-z)^n}{n!\Gamma(b)}\left(\frac{1}{b+n}-\frac{1}{b}\right)\\
=\frac1b-\sum_{n=1}^{\infty}\frac{b\Gamma(b+n)(-z)^n}{n(b+n)n!\Gamma(b)}.
$$
Edit: Ninja'd by 2 minute. 
