Vtyjkyuk

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?
$$ln(1+x) = largeG_2,2^1,2left( x left| beginarraycc 1,1 \ 1,0 endarray right. right).$$
I started but I could not continue...

Let

$a_1=a_2=1$, $b_1=1,b_2=0$, then
beginalign
largeG_2,2^1,2left( x left| beginarraycc 1,1 \ 1,0 endarray right. right)=frac12pi iint_^frac
Gamma(1-s)Gamma(s)^2
Gamma(1+s)prod_3^2(a_j-s)x^s ds
endalign
How is this possible $$prod_3^2(a_j-s)$$.

Also I apply:
beginaligne^-x &= largeG_0,1^1,0left( -x left| beginarraycc - \ 0 endarray right. right)\
&=frac12pi iint_^frac
Gamma(-s)
prod_2^1(1-s)(-x)^s ds
endalign
How $$Gamma(-s)$$ and $$prod_2^1(1-s) $$ are possible?

Thanks.

We have to use certain conventions for these cases.
$$
prod_j=3^2 w_j = 1,quadtextknown as an "empty product"
$$
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
$$
Gamma(z+1) = zGamma(z)
$$
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 $z=-s$:
$$
Gamma((-s)+1) = (-s)Gamma(-s),
\
Gamma(-s) = -fracGamma(1-s)s
$$

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.