Limit involving a hypergeometric function


OK, let's start with an integral representation of that hypergeometric:

$$_2F_1\left(-m,-n,-(m+n);1-\frac{r}{z}\right) \\= \frac{1}{B(-n,-m)} \int_0^1 dx \: x^{-(n+1)} (1-x)^{-(m+1)} \left[1-\left(1-\frac{r}{z}\right)x\right]^m$$

where $B$ is the beta function. Please do not concern yourself with poles involved in gamma functions of negative numbers for now: I will address this below.

As $z \rightarrow 0^+$, we find that

$$\begin{align}\lim_{z \rightarrow 0^+} z^m\ _2F_1\left(-m,-n,-(m+n);1-\frac{r}{z}\right) &= \frac{r^m}{B(-n,-m)} \int_0^1 dx \: x^{-(n-m+1)} (1-x)^{-(m+1)}\\ &= r^m \frac{B(-(n-m),-m)}{B(-n,-m)}\\ &= r^m \lim_{\epsilon \rightarrow 0^+} \frac{\Gamma(-n+m+\epsilon) \Gamma(-n-m+\epsilon)}{\Gamma(-n+\epsilon)^2} \end{align} $$

Note that, in that last line, I used the definition of the Beta function, along with a cautionary treatment of the Gamma function near negative integers, which are poles. (I am assuming that $n>m$.) The nice thing is that we have ratios of these Gamma function values, so the singularities will cancel and leave us with something useful.

I use the following property of the Gamma function (see Equation (41) of this reference):

$$\Gamma(x) \Gamma(-x) = -\frac{\pi}{x \sin{\pi x}}$$

Also note that, for small $\epsilon$

$$\sin{\pi (n-\epsilon)} \approx (-1)^{n+1} \pi \epsilon$$

Putting this all together (I leave the algebra as an exercise for the reader), I get that

$$\lim_{z \rightarrow 0^+} z^m \ _2F_1\left(-m,-n,-(m+n);1-\frac{r}{z}\right) = r^m \frac{n!^2}{(n+m)! (n-m)!}$$

which I believe is equivalent to the stated result.

That last statement is readily seen from writing out the product above:

$$\prod_{i=1}^m \frac{n-(i-1)}{n+1} = \frac{n}{n+1} \frac{n-1}{n+2}\frac{n-2}{n+3}\ldots\frac{n-(m-1)}{n+m}$$

The numerator of the above product is


and the denominator is


The result follows.

Author by


Updated on August 01, 2022


  • M.B.M.
    M.B.M. over 1 year

    I am new to hypergeometric function and am interested in evaluating the following limit:

    $$L(m,n,r)=\lim_{x\rightarrow 0^+} x^m\times {}_2F_1\left(-m,-n,-(m+n);1-\frac{r}{x}\right)$$

    where $n$ and $m$ are non-negative integers, and $r$ is a positive real constant.

    However, I don't know where to start. I did have Wolfram Mathematica symbolically evaluate this limit for various values of $m$, and the patters seems to suggest the following expression for $L(m,n,r)$:


    which one can re-write using the Pochhammer symbol notation as follows:


    If the above is in fact correct, I am interested in learning how to derive it using "first principles" as opposed to the black box that is Wolfram Mathematica. I am really confused by the definition of hypergeometric function ${}_2F_1(a,b,c;z)$, as the defition that uses the Pochhammer symbol in the wikipedia page excludes the case that I have where $c$ is a non-positive integer. Any help would be appreciated.

  • M.B.M.
    M.B.M. over 10 years
    Thanks for the derivation! I have a question: in your first step you bring $z^m$ into the integral and multiply $(1-rx/z)^m$ by it. This results in $(z-rx)^m$, which goes to $(-rx)^m$ as $z\rightarrow 0$ and $r^m$ comes out of the integral. What happens to the negative sign in $(-x)^m$ in that step then? Perhaps I am misunderstanding the Beta function...
  • Ron Gordon
    Ron Gordon over 10 years
    @M.B.M.: thanks for the catch! I made a mistake in writing down the integral representation, so it all works out.
  • M.B.M.
    M.B.M. over 10 years
    @Ror Gordon: no problem! Another question: what happens when $n=m$? I think your approach still works, unless one needs to treat the pole of the gamma function at zero differently from its poles at negative numbers.
  • Ron Gordon
    Ron Gordon over 10 years
    @M.B.M.: I think it works out OK; the pole at $0$ in this case is the same as the others. Just take the limit as $\epsilon \rightarrow 0$ through positive values.
  • M.B.M.
    M.B.M. over 10 years
    It turns out that I used the Pochhammer notation incorrectly in the last line of my question, so I fixed that. Thanks for your help!
  • M.B.M.
    M.B.M. about 10 years
    just want to draw your attention to a related question I posted not long ago. Any help would be greatly appreciated... Thanks!