Appendix: Hypergeometric Identities
We work with the standard definition for the hypergeometric function,
$$\displaystyle \begin{aligned} \,{{}_2}F_1(a,b,c; z) = \sum_{n = 0}^\infty\frac{(a)_n(b)_n}{(c)_n} \frac{z^n}{n!}, \qquad |z|<1. \end{aligned}$$
Of the many identities for hypergeometric functions, we need the following:
for \(c-a-b \notin \mathbb {Z}.\) Hence, thanks to continuity,
We will need to apply a similar identity to (71) but for the setting that c − a − b = 0, which violates the assumption behind (71). In that case, we need to appeal to the formula
$$\displaystyle \begin{aligned} \begin{array}{rcl} {{}_2}F_1 (a,b,a+b,z) & =&\displaystyle \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} \Big( \sum_{k=0}^\infty \frac{(a)_k (b)_k}{(k!)^2} (2\psi(k+1)-\psi(a+k)-\psi(b+k)) (1-z)^k \notag \\ & -&\displaystyle \log(1-z)\, {{}_2}F_1 (a,b,1,1-z)\Big), \end{array} \end{aligned} $$
(73)
for |1 − z| < 1 where the di-gamma function ψ(z) = Γ′(z)∕ Γ(z) is defined for all \(z \ne -n, n \in \mathbb {N}\).
Again, thanks to continuity, we can write
A second identity that is needed is the following combination formula, which states that for any |z| < 1, we have
Its proof can be found, for example at [1]. In the main body of the text, we use this identity for the setting that a = α∕2, b = α and c = 1 + α∕2. This gives us the identity
$$\displaystyle \begin{aligned} {{}_2}F_1\left(\frac{\alpha}{2},\alpha;1+\frac{\alpha}{2};z\right) &= \frac{\Gamma(\alpha/2)\Gamma((2+\alpha)/2)}{ \Gamma(\alpha)} (-z)^{-\alpha/2} {{}_2}F_1\left(\alpha/2,0;1-\alpha/2;\frac{1}{z}\right) \notag \\ &\qquad \qquad + \frac{\Gamma(-\alpha/2)\Gamma((2+\alpha)/2)}{\Gamma((2-\alpha)/2)\Gamma(\alpha/2)} (-z)^{-\alpha} {{}_2}F_1\left(\alpha/2,\alpha;1+\alpha/2;\frac{1}{z}\right) \notag\\ &= \frac{\Gamma(\alpha/2)\Gamma((2+\alpha)/2)}{ \Gamma(\alpha)} (-z)^{-\alpha/2} \notag \\ &\qquad \qquad - (-z)^{-\alpha} {{}_2}F_1\left(\alpha/2,\alpha;1+\alpha/2;\frac{1}{z}\right) , \notag \end{aligned} $$
where we have used the recursion formula for gamma functions twice in the final equality. This allows us to come to rest at the following useful identity
$$\displaystyle \begin{aligned} (-{z})^{-\alpha/2} {{}_2}F_1\left(\alpha/2,\alpha;1+\alpha/2;\frac{1}{z}\right) + (-z)^{\alpha/2} {{}_2}F_1\left(\frac{\alpha}{2},\alpha;1+\frac{\alpha}{2};z\right) &= \frac{\Gamma(\alpha/2)\Gamma((2+\alpha)/2)}{ \Gamma(\alpha)}. {} \end{aligned} $$
(76)
We are also interested in integral formulae, for which the hypergeometric function is used to evaluate an integral. The first is aversion of formula 3.665(2) in [15] which states that, for any 0 < |a| < r and ν > 0, as
$$\displaystyle \begin{aligned} \int_0^{\pi} \frac{\sin^{d-2} \phi }{(a^2+2a r \cos \phi + r^2)^\nu} \mathrm{d}\phi= \frac{1}{r^{2\nu}} B\Big(\frac{d-1}{2}, \frac{1}{2}\Big) \,{{}_2}F_1 \Big(\nu, \nu-\frac{d}{2}+1; \frac{d}{2}; \frac{a^2}{r^2}\Big) , {} \end{aligned} $$
(77)
where B(a, b) = Γ(a) Γ(b)∕ Γ(a + b) is the Beta function. The second is formula 3.197.8 in [15], which states that, for Re(μ) > 0, Re(ν) > 0 and \(|\arg ({u}/{\beta })|<\pi \),
$$\displaystyle \begin{aligned} \int_0^u x^{\nu-1}(u-x)^{\mu-1} (x+\beta)^\lambda \mathrm{d} x = \beta^\lambda u^{\mu+\nu-1} B(\mu,\nu) {{}_2}F_1 \left(-\lambda,\nu;\mu+\nu;-\frac{u}{\beta}\right). {} \end{aligned} $$
(78)
The third is 3.194.1 of [15] and states that, for \(|\arg (1+\beta u)|>\pi \) and Re(μ) > 0, Re(ν) > 0,
$$\displaystyle \begin{aligned} \int_0^u x^{\mu -1} (1+\beta x)^{-\nu}\mathrm{d} x = \frac{u^\mu}{\mu}{{}_2}F_1(\nu, \nu-\mu; 1+\mu; -\beta u), {} \end{aligned} $$
(79)
where 2
F
1 in the above identity is understood as its analytic extension in the event that |βu| > 1.