Hypergeometric function argument simplification Let $_2 F_1 (a,b,c,z)$ be the hypergeometric function. As a result of some integration, I obtained the following expression
$$
f(x) = \frac{\Gamma(2k)\Gamma(2m)}{\Gamma(m)^2\Gamma(k)^2} \frac{\Gamma(m+k)^2}{\Gamma(2k+2m)} \frac{_2 F_1 (2k,k+m,2k+2m,1-\frac{1}{x})}{x^{k+1}}.
$$
The parameters $m$ and $k$ are positive numbers. I will have to use this result in some other integrals, namely the average bit error rate and the capacity integrals:
$$
P(E) = \int_0^{+\infty} \alpha \mathrm{erfc}(\sqrt{\beta x}) f(x) dx,
$$
and
$$
C = \frac{1}{\ln(2)}\int_0^{+\infty} \log(1+x) f(x) dx.
$$
I thought of using the Meijer G function to obtained some closed-form expression but I need to transform the argument of the hypergeometric function in $f(x)$ to be linear. Is there any relation that could help me ? and is there a relation that can simplify the product of such hypergometric function and rational function $\frac{1}{x^{k+1}}$ ?
Thank you.
 A: Here is a transformation for the $_2F_1 (a,b,c,z)$ part, that results in a
form where the argument is linear in $x.\,$ With $z=1-\frac{1}{x}$ we have $\frac{z}{z-1} = 1-x$, and 
$1-z = \frac{1}{x}$. Therefore the first transformation from http://dlmf.nist.gov/15.8.E1 gives 
$$_2F_1 \left(a,b,c,1-\frac{1}{x}\right) = \, _2F_1(a,b,c,z)\\ 
= \,(1-z)^a\, _2F_1 \left(a,c-b,c,\frac{z}{z-1}\right)\\
= x^{-a} \,_2F_1 (a,c-b,c,1-x)
$$
And for the second formula:
$$_2F_1 \left(a,b,c,1-\frac{1}{x}\right) = x^{-b} \,_2F_1 (c-a,b,c,1-x)
$$
You have to attention to range restrictions for $x$, but I think they are valid for $x>0$.
A: I finally got what I was looking for. In line 21 of Table 8.4.49 in "Integrals and Series, vol. 3" by A.P. Prudnikov, Yu.A. Brychkov and O.I. Marichev, we can find a relation between the $_2F_1$ hypergeometric function and the Meijer G function as:
$$
_2F_1 \left(a,b;c;1-\frac{1}{x}\right) = B_{11} G_{22}^{22} \left( x \left| \begin{array}{c} 
1,1+a+b-c \\ a,b \end{array} \right. \right).
$$
By using also a Meijer G representation for $\mathrm{erfc}$ or $\log$, we can solve the above integrals mentioned in the question.
A: Those things are not really very hard only quite tedious since it takes a lot of time to solve them even when using Mathematica, We want to compute;
\begin{equation}
lhs_{k,m} := F_{2,1}\left[ \begin{array}{cc} 2k & k+m \\ 2k+2 m \end{array}; 1- \frac{1}{x} \right]
\end{equation}
Firstly take $m=0$ then:
\begin{eqnarray}
lhs_{k,0}&=&\sum\limits_{p=0}^\infty (k+m)^{(p)} \frac{(1-\frac{1}{x})^p}{p!}\\
&=& \sum\limits_{p=0}^\infty \binom{k+m+p-1}{p} (1-\frac{1}{x})^p \\
&=& \sum\limits_{p=0}^\infty \binom{-k-m}{p} (\frac{1}{x}-1)^p = (1+\frac{1}{x} - 1)^{-k-m}=x^{k+m}
\end{eqnarray}
Now we take $m=1$.
\begin{eqnarray}
lhs_{k,1}&=& \sum\limits_{p=0}^\infty \frac{(2k)(2k+1)}{(2k+p)(2k+p+1)} \cdot (k+m)^{(p)} \frac{(1-\frac{1}{x})^p}{p!}\\
&=& (2k)(2k+1) \sum\limits_{p=0}^\infty \left( \frac{1}{2k+p} - \frac{1}{2k+p+1} \right) \binom{-k-1}{p} (\frac{1}{x} - 1)^p \\
&=& (2k)^{(2)} \int\limits_0^1 \left( t^{2k-1} - t^{2 k} \right) (1+t (\frac{1}{x} - 1))^{-k-1} dt \\
&=& (2k)^{(2)} \frac{x^{2 k}}{(1-x)^{2k+1}} \int\limits_1^{1/x} \frac{(u-1)^{2k-1}}{u^{k+1}} (1-x u) du \\
&=&(2k)^{(2)} \frac{x^{2 k}}{(1-x)^{2k+1}} \sum\limits_{p=0}^{2k-1} \binom{2k-1}{p} (-1)^{p+1}\cdot \\
&&
\left(
\frac{(\frac{1}{x})^{p-k} - 1}{p-k} 1_{k\neq p} + \log(1/x) 1_{k=p} - x \frac{(\frac{1}{x})^{p-k+1} - 1}{p-k+1} 1_{k \neq p+1} - x \log(1/x) 1_{k=p+1}
\right)
\end{eqnarray}
From the above it is not hard to see that those calculations can be carried through to higher values of $m$ and then the generic result can be proven by induction. The result reads:
\begin{eqnarray}
lhs_{k,m}= \frac{(2k)^{(2m)}}{(2m-1)!} \sum\limits_{q=0}^{2m-1} \binom{2m-1}{q} (-1)^q \frac{x^{2k+q}}{(1-x)^{2k+q}} \sum\limits_{p=0}^{2k+q-1} \binom{2k+q-1}{p} (-1)^{q-1-p} 
\left(
\frac{(\frac{1}{x})^{p-k-m+1} -1}{p-k-m+1} 1_{p-k-m \neq 1} + \log(\frac{1}{x}) 1_{p-k-m=1}
\right)
\end{eqnarray}
for $k=1,2,\cdots$ and $m=1,2,\cdots$ and $lhs_{0,m}=1$ and $lhs_{k,0}=x^k$.
    In[18]:= mmax = 5;
mat = Table[ 
   Hypergeometric2F1[2 k, k + m, 2 k + 2 m, 1 - 1/x], {k, 0, 
    mmax}, {m, 0, mmax}];
tmp = Simplify[
   Table[Pochhammer[2 k, 2 m]/(2 m - 1)! Sum[ 
       Binomial[2 m - 1, q] (-1)^q x^(2 k + q)/(1 - x)^(2 k + q) Sum[
         Binomial[2 k + q - 1, p] (-1)^(q - 1 - p) If[
           p - k - m != -1, (1/x^(p - k - m + 1) - 1)/(p - k - m + 1),
            Log[1/x]], {p, 0, 2 k + q - 1}], {q, 0, 2 m - 1}], {k, 0, 
      mmax}, {m, 0, mmax}] - mat, Assumptions -> x > 1];
MatrixForm[tmp]


