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The following expression involving a hypergeometric function popped out as a solution to an integral that I've been working on (via Gradshteyn and Ryzhik):

$$L(a,b)=\lim_{z\rightarrow1^-}{}_2F_1(-a,-b,-(a+b),z)$$

where $a\geq0$ and $b\geq0$ are integers and limit is taken from below. According to the Wikipedia article, it is undefined or infinite for the negative third term (which is my case). However, Wolfram Mathematica happily evaluates not just the limit but the expression ${}_2F_1(-a,-b,-(a+b),1)$ for seemingly all $a$ and $b$ fitting my criteria (for example, according to Mathematica, ${}_2F_1(-3,-5,-8,1)=\frac{1}{56}$ and ${}_2F_1(-4,-6,-8,1)=\frac{1}{210}$).

Now there are expressions for the values of this hypergeometric function at argument unity on DLMF, but they don't apply in my case since the third argument is negative.

What is going on here? Is there an expression for $L(a,b)$ in my case? Any help/pointers would be appreciated.

This is sort of related to my previous question, though I don't think the technique in that answer works here (the integral in the representation involving the beta function doesn't converge).

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    $\begingroup$ Are $a$ and $b$ always integers? In that case ${}_2F_1(-a,-b,-(a+b),z)$ is a polynomial of degree $c = \min\{a,b\}$. The product of Pochhammer symbols $(a)_n(b)_n$ evaluates to $0$ for all $n > c$, so the series defining the hypergeometric function terminates there. For example, $${}_2F_1(-3,-5,-8,z) = 1-\frac{15}{8}z+\frac{15}{14}z^2-\frac{5}{28}z^3.$$ In that case, $${}_2F_1(-3,-5,-8,1) = \sum_{n=0}^{3} \frac{(-3)_n(-5)_n}{(-8)_n n!}.$$ $\endgroup$ – Antonio Vargas Sep 27 '13 at 1:07
  • $\begingroup$ Yes, $a$ and $b$ are always integers. This is helpful (though looks like I got an alternating progression here, which is a pain)... $\endgroup$ – M.B.M. Sep 27 '13 at 1:20
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Using the Chu-Vandermonde identity, we can recast the limit in a computationally convenient form:

$$\lim_{z\rightarrow1^-}{}_2F_1\left({{-a,\quad-b}\atop{-a-b}}\middle|z\right)=\frac{(\max(a,b)-a-b)_{\min(a,b)}}{(-a-b)_{\min(a,b)}}$$

Since the function is symmetric in $a$ and $b$, we choose the larger of $a$ and $b$ in the argument and the smaller one in the index, to keep the arguments as small as possible.

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If a,b are positive integers then the function is a polynomial. Look at A&S (the original Handbook of Mathematical Functions) section 15.1 : http://people.math.sfu.ca/~cbm/aands/abramowitz_and_stegun.pdf Which is apparently legitimate from SFU.

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