2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $I_n$.

$\endgroup$
0
$\begingroup$

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: $x, 1-x$...

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.