Help with Mellin-Barnes Integral (product of two Hypergeometrics)

Question

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

Answer 1

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!!

Answer 2

The hypergeometric function in contour form is $${}_{2}F_{1}(a, b; c; x) = \frac{1}{2 \pi i} \, \int_{-i \infty}^{i \infty} \frac{\Gamma(-s) \Gamma(a+s) \Gamma(b+s)}{\Gamma(c + s)} \, ds.$$ The integral in question becomes: \begin{align} I &= \int_{0}^{1} {}_{2}F_{1}(p,p;2p;t) \cdot {}_{2}F_{1}\left(\frac{1+2a}{2},\frac{1-2a}{2};1;\frac{t-1}{t}\right) \, t^{p-2} \, dt \\ &= \left(\frac{1}{2 \pi i}\right)^2 \, \int \int \frac{\Gamma(-u) \Gamma^2(p+u)}{\Gamma(2p + u)} \cdot \frac{\Gamma(-s) \Gamma(\frac{1}{2} + a + s) \Gamma(\frac{1}{2} - a + s)}{\Gamma(s + 1)} \, J \, du \, ds, \end{align} where \begin{align} J &=\int_{0}^{1} t^{p-2} \, (-t)^{u} \, \left(-1 + \frac{1}{t}\right)^{s} \, dt = (-i)^{u} \, B(s+1, p+u-s-1) \\ &= (-1)^{u} \, \frac{\Gamma(s+1) \Gamma(p+u-s-1)}{\Gamma(p+u)}. \end{align} Now, by use of Barnes' Lemma (first) given by $$\frac{1}{2 \pi i} \, \int_{-i \infty}^{i \infty} \Gamma(a+s) \Gamma(b+s) \Gamma(c -s) \Gamma(d -s) \, ds = \frac{\Gamma(a+c) \Gamma(a+d) \Gamma(b+c) \Gamma(b+d)}{\Gamma(a+b+c+d)},$$ with $d = 0$, $c = p+u-1$, $a = 1/2 +a$ and $b = 1/2 -a$, then $$ \frac{1}{2 \pi i} \, \int_{- i \infty}^{i \infty} \Gamma(-s) \Gamma(\frac{1}{2} + a + s) \Gamma(\frac{1}{2} - a + s) \Gamma(p+u-s-1) \, ds = \frac{\Gamma(a + \frac{1}{2}) \Gamma(p+a-\frac{1}{2} + u) \Gamma(p - a - \frac{1}{2} +u)}{\Gamma(p+u)} $$ and \begin{align} I &= \left(\frac{1}{2 \pi i}\right) \, \int \frac{(-1)^u \Gamma(-u) \Gamma(p+u)}{\Gamma(2p + u)} \, du \\ & \hspace{5mm} \cdot \frac{1}{2 \pi i} \, \int_{- i \infty}^{i \infty} \Gamma(-s) \Gamma(\frac{1}{2} + a + s) \Gamma(\frac{1}{2} - a + s) \Gamma(p+u-s-1) \, ds \\ &= \frac{\Gamma(a + \frac{1}{2})}{2 \pi i} \, \int_{- i \infty}^{i \infty} \frac{\Gamma(-u) \Gamma(p-a-\frac{1}{2} + u) \Gamma(p +a-\frac{1}{2} +u)}{\Gamma(2 p + u)} \, (-1)^{u} \, du \\ &= \Gamma\left(a + \frac{1}{2}\right) \, {}_{2}F_{1}\left(p+a-\frac{1}{2}, p-a-\frac{1}{2}; 2p ; 1\right) \\ &= \frac{\Gamma(2p) \Gamma\left(a+\frac{1}{2}\right)}{\Gamma\left(p+a+\frac{1}{2}\right) \Gamma\left(p-a+\frac{1}{2}\right)}. \end{align}