Integration of quotient of hypergeometric functions If one has an expression of the form,
$$\int \frac{\, _2F_1(a,b;c;z)}{_2F_1(d,e;f;z)}dz,$$
where the arguments are all complex numbers, is it possible to integrate this to get another hypergeometric function, or some other analytic expression?
If one expands the hypergeometric functions into their series representation, it seems straightforward that this expression can be integrated, however possibly the resulting series is no longer convergent. When I tried to evaluate this integral with Mathematica, it won't give me an output, so does this mean that the new series is non-convergent? Or is there some other way I can analyse the series?
In particular, the integral is,
$$\int \frac{\, _2F_1((3 + \alpha),  (3 - \sqrt{\beta} + \alpha); 3 + 
 \alpha; z)}{_2F_1( (1 + \alpha),  (1 - \sqrt{\beta} + \alpha ); 1 + 
 \alpha; z)}dz,$$
where $\alpha$ and $\beta$ are constants.
Thanks!
 A: The ratio of two analytic functions is analytic on any domain where the denominator is not $0$. Also the integral of a power series has the same radius of congergence as the original power series (as can be easily verified by the ratio test).
So yes, this integral exists on any simply-connected domain that does not include a zero of the denominator, and the integral will also be analytic over that domain. However, it is not likely to be another hypergeometric function.
A: The integration of the quotient of hypergeometric functions can be calculated or verified with the help of a CAS such as Mathematica.
The Gaussian hypergeometric function is defined by:
$\displaystyle \, _2F_1(a,b;c;z)=\sum _{k=0}^{\infty } \frac{z^k (a)_k (b)_k}{k! (c)_k},$
where the Pochhammer symbol is defined by
$\displaystyle x_{(n)}=x(x+1)(x+2)\cdots (x+n-1)=\prod _{k=1}^{n}(x+k-1)=\prod _{k=0}^{n-1}(x+k)$
${\displaystyle x_{(n)}={\frac {\Gamma (x+n)}{\Gamma (x)}}}$
The quotient of the two hypergeometric functions in this question is (verified with Mathematica):
$\displaystyle \frac{\, _2F_1\left(\alpha +3,\alpha -\sqrt{\beta }+3;\alpha +3;z\right)}{\, _2F_1\left(\alpha +1,\alpha -\sqrt{\beta }+1;\alpha +1;z\right)}=\frac{1}{(1-z)^2}$
Hence the integral is given by
$ \displaystyle \int \frac{\, _2F_1\left(\alpha +3,\alpha -\sqrt{\beta }+3;\alpha +3;z\right)}{\, _2F_1\left(\alpha +1,\alpha -\sqrt{\beta }+1;\alpha +1;z\right)} \, dz=\int \frac{1}{(1-z)^2} \, dz=\frac{1}{1-z}+C$
For arbitrary numbers $p$ and $q,$ the integral of the quotient of hypergeometric functions is:
$\displaystyle \int \frac{\, _2F_1\left(p+\alpha ,p+\alpha -\sqrt{\beta };p+\alpha ;z\right)}{\, _2F_1\left(q+\alpha ,q+\alpha -\sqrt{\beta };q+\alpha ;z\right)} \, dz=\int (1-z)^{q-p} \, dz=\frac{(1-z)^{-p+q+1}}{p-q-1}+C$
