Help with Mellin-Barnes Integral (product of two Hypergeometrics) I am trying to prove that
$$\int_0^1 \frac{dz}{z^2} z^{h}\cdot {}_{2}F_{1}(h,h;2h;z) \cdot {}_{2}F_{1}\left(\frac{1+2a}{2},\frac{1-2a}{2};1;\frac{z-1}{z}\right) = -\frac{\Gamma(2h)}{\Gamma{(h)}^2} \cdot \left(\frac{1}{a^2-(h-1/2)^2} \right).$$
To begin I first use Mellin-Barnes representation of Hypergeometric representation:
$$ {}_{2}F_{1}(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\int_{-i\infty}^{i\infty}ds\frac{\Gamma(-s)\Gamma(a+s)\Gamma(b+s)}{\Gamma(c+s)}(-z)^s$$
The integral over the z variable is (done in Mathematica):
$$\int_0^1 \frac{z^h(-z)^s(1-z)^t}{z^2z^t}=(-1)^s\frac{\Gamma(-1+h+s-t)\Gamma(1+t)}{\Gamma(h+s)} $$
Plugging this back in I get,
$$\frac{\Gamma(2h)}{\Gamma(h)^2\Gamma(1/2+a)\Gamma(1/2-a)}\int\int ds dt \frac{\Gamma(-1+h+s-t)\Gamma(-s)\Gamma(h+s)\Gamma(-t)\Gamma(1/2+a+t)\Gamma(1/2-a+t)}{\Gamma(2h+s)}(-1)^s$$
where s and t are the Mellin-Barnes variables. 
Then I use Barnes lemma to integrate over the t-variable to obtain:
$$\frac{\Gamma(2h)}{\Gamma(h)^2}\int_{-i\infty}^{i\infty}\frac{\Gamma(-s)\Gamma(-1/2+h+s+a)\Gamma(-1/2+h+s-a)}{\Gamma(2h+s)} \, ds$$
I dont see how I will get to the final result from here.
Any help appreciated.
Thanks

I just figured it out.
The last integral can be written in terms of 2F1 Hypergeometric using standard Mellin-Barnes integral.
And then I needed to use the Gauss Hypergeometric formula.
Yay!!
