Can any one prove for me $\ln(1+x) = \large{G}_{2,2}^{1,2}\left( x \left| \begin{array}{cc} 1,1 \\ 1,0 \end{array} \right. \right).$ I am a PhD student in Wireless Communications and recently I found a paper about the use of "The generalized upper incomplete Fox’s H function". 
I think that in order to understand this function, I need before to understand the Meijer’s G-Function. The thing is I do not understand how to use this formula of Meijer’s G-Function.
Could anyone help me how to prove this formula?
 $$\ln(1+x) = \large{G}_{2,2}^{1,2}\left( x \left| \begin{array}{cc} 1,1 \\ 1,0 \end{array} \right. \right).$$
I started but I could not continue...
Let
$a_1=a_2=1$, $b_1=1,b_2=0$, then
\begin{align}
\large{G}_{2,2}^{1,2}\left( x \left| \begin{array}{cc} 1,1 \\ 1,0 \end{array} \right. \right)=\frac{1}{2\pi i}\int_{}^{}\frac
{\Gamma(1-s)\Gamma(s)^2}
{\Gamma(1+s)\prod_{3}^{2}(a_j-s)}x^s ds
\end{align}
How is this possible $$\prod_{3}^{2}(a_j-s)$$.
Also I apply: 
\begin{align}e^{-x} &= \large{G}_{0,1}^{1,0}\left( -x \left| \begin{array}{cc} - \\ 0 \end{array} \right. \right)\\
&=\frac{1}{2\pi i}\int_{}^{}\frac
{\Gamma(-s)}
{\prod_{2}^{1}(1-s)}(-x)^s ds
\end{align}
How $$\Gamma(-s)$$ and $$\prod_{2}^{1}(1-s) $$ are possible?
Thanks.

We have to use certain conventions for these cases.
$$
\prod_{j=3}^2 w_j = 1,\quad\text{known as an "empty product"}
$$
and similarly
$$
\prod_{j=2}^1 w_j = 1.
$$
The $\Gamma$ function is defined by an integral for positive arguments, but may be extended to other arguments.  The functional equation
$$
\Gamma(z+1) = z\Gamma(z)
$$
is used for that.  When $-1<z<0$, we have $z+1$ where $\Gamma(z)$ is known.  So for $0<s<1$, apply this with $z=-s$:
$$
\Gamma((-s)+1) = (-s)\Gamma(-s),
\\
\Gamma(-s) = -\frac{\Gamma(1-s)}{s}
$$
I note that you have not specified the integraion path here; it is a path in the complex plane, and you will need $\Gamma(-s)$ on that path.
Note
It may be an interesting exercise to evaluate the function this way.  But I think in practice we would use the differential equation to do this.
