Poles of hypergeometric function $_2F_1$

Question

We consider the hypergeometric function $ _2F_1 [\dfrac{1}{2}(1+k+l+\omega), \dfrac{1}{2}(1+k-l+\omega), 1+k, -r^2]$, and use its expansion as given in [1] in terms of the rising Pochhammer symbols. Here $k$ is a positive real, $l$ is a positive integer and $\omega$ is a real. So we can say that the poles of the above function are given by $\omega = -2n - l - k - 1$ and $\omega = -2n + l - k - 1$ in the $\omega$ plane, using the same expansion given in [1], as these are the points at which the Gamma functions in the rising Pochhammer symbols blow up. Here $n$ is a positive integer. Is it correct upto this stage?

Now consider Eqn[4.1.5] of [2], which is the same function I have written above with factors of r in front. If the previous step is correct, then why

1. are the poles of the function as in the Eqn[4.1.8] of [2] different from the above-mentioned one
2. how do we obtain Eqn[4.1.6] and Eqn[4.1.7] in limit $r \rightarrow \infty$, and how does one get Eqn[4.1.8] from them?

[1] - https://en.wikipedia.org/wiki/Hypergeometric_function#The_hypergeometric_series

[2] - https://arxiv.org/pdf/0812.2909.pdf

Answer 1

So you are analizing the Hypergeometric function $$ \bbox[lightyellow] { \eqalign{ & {}_2F_{\,1} \left( {1/2\left( {1 + k + l + \omega } \right),1/2\left( {1 + k - l + \omega } \right);\;k + 1;\; - r^{\,2} \,} \right) = \cr & = \sum\limits_{0\, \le \,j} {{{\left( {1/2\left( {1 + k + l + \omega } \right)} \right)^{\,\overline {\,j\,} } \left( {1/2\left( {1 + k - l + \omega } \right)} \right)^{\,\overline {\,j\,} } } \over {\left( {k + 1} \right)^{\,\overline {\,j\,} } 1^{\,\overline {\,j\,} } }}\left( { - r^{\,2} } \right)^{\,j} } \cr} } \tag{1}$$ where

• ${x^{\,\overline {\,j\,} } }$ denotes the Rising Factorial, i.e. the (rising) Pochammer symbol, and so ${1^{\,\overline {\,j\,} } }=j!$

• $0<k\in \mathbb{R}$, $ l \in \mathbb{N}$, $\omega \in \mathbb{R}$

and you are concerned about its poles, in terms of the variable parameter $\omega$.

On this subject, there are various points to pay attention to and clarify:

1. The above series expression for the Hypergeometric function is the standard expression valid in general for $z$, here $(-r^2)$, inside the unit circle $|z|<1$.
2. For $1 \le |z|$ , and so for $1 \le |r|$, the hypergeometric function shall be computed by analytical continuation of the expression above (re. e.g. to this Mathworld article).
3. Because of the alternating sign introduced by $(-r^2)^j$ it might be possible that for special values of the parameters the sum in (1) be convergent for any value of $r$, but we are not going to investigate this special case.
4. For $ |r| < 1$, the expression in (1) does not have any finite pole in $\omega$ .
   Consider in fact that the rising factorial, for non-negative integer exponent (as it is the $j$ in the sum), is defined as $$ \bbox[lightyellow] { z^{\,\overline {\,n\,} } \quad \left| {\;0 \le {\rm integer}\,n} \right.\quad = \prod\limits_{0\, \le \,k\, \le \,n - 1\,} {\left( {z + k} \right)} = {{\Gamma \left( {z + n} \right)} \over {\Gamma \left( z \right)}} }$$ which is a polynomial in $z$ of degree $n$ defined in all $\mathbb C$, and thus has $n$ zeros and no finite pole.
   The definition of the rising factorial through the ratio of the Gamma, is to be taken in the limit when approaching a pole of the Gamma in the numerator, which - if $n$ is non-negative - cancels with the pole at the denominator, leaving the ratio of the residuals, since both poles are simple.
   That is $$ \bbox[lightyellow] { \left( { - m} \right)^{\,\overline {\,n\,} } \quad \left| {\;0 \le {\rm integers}\,n,m} \right.\quad = \prod\limits_{0\, \le \,k\, \le \,n - 1\,} {\left( {k - m} \right)} = \mathop {\lim }\limits_{z\; \to \; - m} {{\Gamma \left( {z + n} \right)} \over {\Gamma \left( z \right)}} }$$

Having cleared the above points, then the expression provided in [4.1.6] of your ref. [2], is said to be "for large$r$", thus it should involve an asymptotic expansion of the hypergeometric, which cannot be that indicated in the Mathworld article cited above because in present case the condition $a-b = l \notin \,\mathbb{Z}$ is not satisfied.

I am not going here to investigate which expansion has been used.

Yet, if that is correct, then in eq. [4.1.7] it is the digamma function inside $\beta(\omega,k,l)$ that has poles, while the function $\alpha(\omega,k,l)$ has no poles, but zeros, which are located differently than the digamma poles. Either the zeros of $\alpha$ and poles of $\beta$ are simple . So in the product $\alpha(\omega,k,l) \beta(\omega,k,l)$ the poles are those given by the digamma which are not cancelled by a zero in $\alpha$.