Appendix: Hypergeometric functions [024C]

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Appendix: Hypergeometric functions

Recall that the hypergeometric function $\,{}_{2}F_{1}[a,b;c;z]$ is defined by

\,_{2}F_{1}[a,b;c;z]=\sum_{n=0}^{\infty}\frac{(a)_{n}(b)_{n}}{n!(c)_{n}}z^{n}

(38)

where

(\alpha)_{n}:=\alpha(\alpha+1)(\alpha+2)\cdots(\alpha+n-1),\quad(\alpha)_{0}=1

and we assume that $c\notin\mathbb{Z}_{\leq 0}$ . It is a standard fact that the hypergeometric series converges when $|z|<1$ and when ${\rm Re}(c-a-b)>0$ is also converges when $z=1$ ; see for instance [1, Chapter 1].

We start from the elementary Taylor expansion

(1-y)^{-\beta}=\sum_{k=0}^{\infty}\frac{(\beta)_{k}}{k!}y^{k},\quad|y|<1.

Lemma 5.2.

For $x\geq 1$ we have

\int_{1}^{x}(1-y^{-2})^{-1/n}dy=x\,_{2}F_{1}[-\frac{1}{2},\frac{1}{n};\frac{1}{2};x^{-2}]-\,_{2}F_{1}[-\frac{1}{2},\frac{1}{n};\frac{1}{2};1].

Proof.

We compute

\displaystyle\int_{1}^{x}(1-y^{-2})^{-1/n}dy=\int_{1}^{x}\sum_{k=0}^{\infty}\frac{(\frac{1}{n})_{k}}{k!}y^{-2k}dy=y\sum_{k=0}^{\infty}\frac{(\frac{1}{n})_{k}}{k!}\frac{y^{-2k}}{(-2k+1)}\bigg|_{y=1}^{y=x}.

The Lemma follows from the observation $\frac{(-\frac{1}{2})_{k}}{(\frac{1}{2})_{k}}=-\frac{1}{2k-1}.$ ∎

Proposition 5.3.

Let $n\geq 3$ . Suppose $g$ satisfies the ODE

\frac{d}{dy}\left(\frac{g}{y}\right)=\frac{1}{y^{2}(1-y^{n})^{\frac{1}{2}}},\qquad 0<y<1,

together with the initial conditions $g(0)=-1,g^{\prime}(0)=0$ . Then we have

g(y)=-\,_{2}F_{1}[\frac{1}{2},-\frac{1}{n};\frac{n-1}{n};y^{n}].

In particular

g(1)=\lim_{y\to 1}g(y)=-\,_{2}F_{1}[\frac{1}{2},-\frac{1}{n};\frac{n-1}{n};1]=-\frac{\Gamma(\frac{n-1}{n})\sqrt{\pi}}{\Gamma(\frac{n-2}{2n})}.

Proof.

Integrating

y^{-2}(1-y^{n})^{-1/2}=\sum_{k=0}^{\infty}\frac{(\frac{1}{2})_{k}}{k!}y^{nk-2},

and utilizing the initial conditions to fix the leading coefficients,

g(y)=\sum_{k=0}^{\infty}\frac{(\frac{1}{2})_{k}}{k!}\frac{y^{nk}}{nk-1}.

The formula for $g(y)$ follows from the observation $\frac{-1}{nk-1}=\frac{(-\frac{1}{n})_{k}}{(\frac{n-1}{n})_{k}}.$ The evaluation of $g(1)$ appeals to Gauss’ hypergeometric theorem [1, Section 1.3]:

Lemma 5.4.

For $a,b,c$ with ${\rm Re}(c-a-b)>0$ we have

\,{}_{2}F_{1}[a,b;c;1]=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}.

∎