A definite integral of hypergeometric function 2F1

Question

I am wondering whether there exists a closed form for the definite integral $$F(x)=\int_0^1t^{-a}(1-t)^{N}(1-xt)^{-a}{}_2F_1(-a,k-a-1/2,k-a;4xt(1-xt))dt,$$ where $a\in(0,1)$ and $N,k$ are positive integers.

It seems that I cannot apply Kummer's quadratic transformation formula directly. So I tried to follow the method of proving Kummer's formula by expanding $_2F_1$, and I require some formula for $_2F_1(A,-A,C;x)$. However I cannot find anything relevant in many literatures.

Do you have some other ideas?

Thanks in advance.

Answer 1

The integral in question is given by,

$$I = \int_{0}^1 t^{-a} (1-t)^N (1-xt)^{-a}F[-a,k-a-1/2,k-a;4xt(1-xt)] \, \mathrm{d}t$$

where $F$ denotes the standard hypergeometric series which is only well-defined for $k-a\geq 0$, and the Pochhammer symbol only terminates if either, $a \geq 0$ or $k-a-1/2 \leq 0.$ Expanding the series yields,

$$F = \sum_{n=0}^\infty \frac{(-a)_{n} (k-a-1/2)_n}{(k-a)_n} \frac{(4xt(1-xt))^n}{n!}$$

where $(q)_n$ denotes the rising Pochhammer symbol (contrary to popular notation). We approach the original integral $I$ by integrating term by term as we slowly expand $F$. The case $n=0$ is,

$$I_{0}=\int_{0}^1 t^{-a}(1-t)^N (1-xt)^{-a} \, \mathrm{d}t$$

as the hypergeometric series is unity for $n=0.$ Notice the integral $I_0$ is precisely an Appell series, i.e.

$$I_0= \frac{\Gamma (1-a)\Gamma(N+1)}{\Gamma(N+a+1)}\mathcal{F}[(1-a),a,0,(N+a+1);x,0]$$

where $\mathcal{F}$ denotes the Appell series, rather than the hypergeometric series. We may proceed similarly for the subsequent $n=1$ contribution, namely, $$I_1 = 4x\frac{a^2(1-a)(k-a-1/2)^2(k-a+1/2)}{(k-a)^2(k-a-1)}\int_0^1 t^{-a+1}(1-t)^N (1-xt)^{-a+1}$$

Luckily $I_1$ is essentially an Appell series also, namely,

$$I_1 = 4x\frac{a^2(1-a)(k-a-1/2)^2(k-a+1/2)}{(k-a)^2(k-a-1)} \frac{\Gamma(2-a)\Gamma(N+1)}{\Gamma(N+3-a)} \\ \times\mathcal{F}[(2-a),(a-1),0,(N+a+1);x,0]$$

Proceed similarly for $n\geq 2$. Goodluck, it's quite a messy problem! Maybe after several iterations you may deduce a general expression for the $n$th contribution $I_n$.

Answer 2

May be use the series for $_{2}F_{1}$ and then integrate term by term? The result looks like a sort of Appell-type function of two arguments: $x, 1-x$...

Answer 3

The Maple 18 code

A := eval(convert(hypergeom([-a, k-a-1/2], [k-a], y), FPS, y), y = 4*x*t*(-t*x+1)) assuming k::posint:
int(A*t^(-a)*(1-t)^N*(-t*x+1)^(-a), t = 0 .. 1) assuming a>0,a<1,N::posint,x>0,x<1;

outputs the integral under consideration as the series $$\Gamma(N+1)\times$$ $$\sum\limits_{i=0}^\infty \frac{\mathop{\rm pochhammer}(-a,i)x^{i}2^{2i}{\mbox{$_2$F$_1$}(- i+a,1-a+ i;-a+ i+2+N;x)}}{\mathop{\rm pochhammer}(k-a,i)\Gamma(-a+i+2+N)\mathop{\rm pochhammer}(k-a-1/2,i)^{-1}i!},$$ where $\mathop{\rm pochhammer}$ is described here. See here for the Maple output.