Let $n,N,T$ be positive integers, with $N=\binom{n}{2}$, and $3\leq n\leq T\leq N$. Define:

$$P(z):=z^{N+1-T}\int_0^1\frac{(1-t^2)^{n-2}}{(1-(1-z)t)^{N+1}}{}_2F_1\left[-T,N+1,N+2-T; \frac{tz}{1-(1-z)t}\right]dt.$$

I am interested in evaluating the limit $\lim_{z\rightarrow 0}P(z)$. I am wondering if there is any nice expression for this limit, perhaps as another hypergeometric function. I would be thrilled if the limit is another integral. In fact, I know that $P(z)$ itself is a polynomial, so there's a lot of magic happening to make this work.

The difficulty is resolving the singularity at $z=0,t=1$. It seems like all of the terms in the hypergeometric function contribute to this integral. Notice that one cannot just write the hypergeometric function as a sum and then swap the swap the sum with the integral and the limit. I've tried a number of Kummer transformations to somehow make things nicer but I haven't see much progress. I've consulted a number of integral tables but haven't seen such an integral come up.

One thing I've noticed is that since $-T$ is a negative integer, the hypergeometric function is a finite polynomial. In fact, it is proportional to the Jacobi polynomial $P_T^{(N+1-T,-1)}\left[\frac{1-t(1+z)}{1-t(1-z)}\right]$. I mention this because the $(1-t^2)^{n-2}$ term reminds me of the orthogonality weight for certain Jacobi polynomials, so perhaps some nice inner product is being computed here? Moreover, the full ratio $(1-t^2)^{n-2}/(1-(1-z)t)^{N+1}$ looks like the primary term involved in the Schlafli integral representation for Associated Legendre polynomials, although there the integral is on $[-1,1]$ or on a contour surrounding $1/(1-z)$.

I was thinking of splitting the integral into two pieces: $\int_0^1=\int_0^c+\int_c^1$, with hopes that if $c$ is chosen carefully (as a function of $z$), only the latter integral contributes. It baffles me though, how to obtain some kind of uniform bound on the hypergeometric part with $z\approx 0$ and $t\approx 1$. Is there perhaps some modification of steepest descent or saddle point methods that would apply here?

  • $\begingroup$ could you perhaps try using an integral representation of $ _2F_1$ and swap limits of integration (assuming this is legal to do)? $\endgroup$ Commented May 7, 2014 at 23:35

1 Answer 1


Not a full proof, but a "closed" formula with two independent finite sums. I am sure that the proof can be made more compact, but I hadn't have the time to think about it.

Very important: I assume without proof that the resulting $P(z)$ is indeed a polynomial in $z$, which includes in particular, that $P(0)$ exists.

First one performs a transformation of integration variable, $t=\left(1+\frac{z}{y}\right)^{-1}$, to get rid of the $z$ dependence of the hypergeometric, $$ P(z)=z^{-T-1-n}\ 2^{n-2}\int_{0}^{\infty}dy\ y^{-n} \, \left(1+\frac{1}{y}\right)^{-N-1} \,\left(1+\frac{z}{y}\right)^{N-2n+3}\, \left(1+\frac{z}{2y}\right)^{n-2} \\ _{2}F_{1}\left(-T,N+1,N+2-T;\left(1+\frac{1}{y}\right)\right). $$ Note that the exponents of the binomials containing $z$ are positive and integer, and the argument of the hypergeometric is independent of $z$. For convenience one might perform a linear transformation of the hypergeometric (to be found e.g. in DLMF http://dlmf.nist.gov/15.8.i) $$ _{2}F_{1}(a,b,d:x)=(1-x)^{-a} \ _{2}F_{1}(a,c-b,c;\frac{x}{x-1}), $$ and expand the binomials containing $z$: $$ P(z)=z^{-T-1+n}\ 2^{n-2}\int_{0}^{\infty}dy\ y^{-n-T} \, \left(1+\frac{1}{y}\right)^{-N-1-T} \\ _{2}F_{1}(-T,1-T,N+2-T;-y) \\ \sum_{r=0}^{N-2n+3}z^{r}y^{-r}{N-2n+3 \choose r}\, \sum_{s=0}^{n-2}z^{s}2^{-s}y^{-s}{n-2 \choose s} . $$

The hypergeometric is a polynomial (because the first parameters are negative $-T<0$ and $1-T<0$) $$ _{2}F_{1}(-T, 1-T, N+2-T; -y)= \sum_{k=0}^{T-1} \ \frac{{T-1 \choose k}{T \choose k}}{{N+1-T+k \choose k}} \ (-y)^{k} $$ The integrals converge for all considered parameter values and thus summation and integration can be interchanged $$ P(z)=2^{n-2}\ \sum_{r=0}^{N-2n+3}\sum_{s=0}^{n-2}{N-2n+3 \choose 3}2^{-s}{n-2 \choose s} \\ z^{-T-1+n+r+s} \sum_{k=0}^{T-1}\frac{{T-1 \choose k}{T \choose k}}{{N+1-T+k \choose k}} \ (-1)^{k}\ \int_{0}^{\infty}\, dx \ y^{k-r-s-n-T}\, \left(1+\frac{1}{y}\right)^{-N-1-T}. $$ Now it is manifestly evident, that $P(z)$ is a finite sum of powers of $z$. It might be proven that all coefficients of $z^{-M}$ for positive $M$ are vanishing (thus $P(z)$ being a polynomial), but I could not succeed, although the numerical evidence is strong.

It can be easily seen that for all allowed parameter values the integral converges to give $$ \int_{0}^{\infty}\, dx \ y^{M-2T+k-1}\, \left(1+\frac{1}{y}\right)^{-N-1-T} = \frac{(N-T+k+M)!(2T-k-M-1)!}{(N+T)!}, $$ with the abbreviation $$ -M = -T-1+n+r+s $$ thus $$ -N+t\leq M\leq T+1-n. $$ Next one sets $M=0$, which fixes the exponent of $z$ to 0 and collapses the sum over, say, $r$. After a little rearrangement we get $$ P(0) = \frac{2^{n-2}}{(N+T)!} \sum_{s=0}^{n-2} {N-2n+3 \choose T+1-n-s}{n-2 \choose s} 2^{-s} \\ \sum_{k=0}^{T-1} \frac{{T-1 \choose k}{T \choose k}}{{N+1-T+k \choose k}} \ (-1)^{k} \frac{(N-T+k+M)!(2T-k-M-1)!}{(N+T)!}. $$ Mathematica offered more simplifications, but some where plagued with abiguities (like "0 times $\infty$") for certain (otherwise valid) parameter combinations, so I leave these out here.

Edit: However, the some over $k$ is simplified by Mathematica to a generalized hypergeometric function with unity argument $$ \frac{{T-1 \choose k}{T \choose k}}{{N+1-T+k \choose k}} \ (-1)^{k} \frac{(N-T+k+M)!(2T-k-M-1)!}{(N+T)!} = \\ \frac{1}{2T-M}{N+T \choose M+N-T}^{-1}\ _{3}F_{2}(-T,-T+1,M+N-T+1,N-T+2,M-2T+1;1). $$ As it turns out the hypergeometric function is Saalschützian, which allows a substantial simplification using Pochhammer's symbols (see e.g. http://dlmf.nist.gov/16.4.E3) $$ F_{2}(-T,-T+1,M+N-T+1,N-T+2,M-2T+1;1) = \frac{(N+1)_{T} (-M+1)_{T}}{(N-T+2)_{T}(-M+T)_{T}}. $$ While all other terms stay finite in the considered parameter range the term $(-M+1)_{T}$ gives zero for $M>0$. Hence, the coefficients negative powers of of $z$ are vanishing, supporting the assumption that $P(z)$ is indeed a polynomial in $z$.


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