Is there any relations between the following hypergeometric functions? $$\ _2F_1(1,-a,1-a,\frac{1}{1-z})$$ $$\ _2F_1(1,-a,1-a,{1-z})$$ $$\ _2F_1(1,a,1+a,\frac{1}{1-z})$$ $$\ _2F_1(1,a,1+a,{1-z})$$


First, you need to use Barnes integral representation $${}_2F_1(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\frac{1}{2\pi i} \int_{-i\infty}^{+i\infty}\frac{\Gamma(a+s)\Gamma(b+s)\Gamma(-s)}{\Gamma(c+s)}(-z)^sds. $$ The Gauss hypergeometric function ${}_2F_1(a,b;c;z)$ is usually defined by a power series that converges only for $|z|<1$, but you have to extend the definition on the whole complex plane such that the function can be evaluated at both $1-z$ and $1/(1-z)$.

Using the integral representation, \begin{align*} \ _2F_1(1,-a;1-a;{1-z}) &= \frac{\Gamma(1-a)}{\Gamma(1)\Gamma(-a)}\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \frac{\Gamma(1+s)\Gamma(-a+s)\Gamma(-s)}{\Gamma(1-a+s)}(z-1)^sds \cr &=\frac{(-a)}{2\pi i}\int_{-i\infty}^{+i\infty}\frac{\Gamma(1+s)\Gamma(-s)}{s-a}(z-1)^sds, \end{align*} and \begin{align*} \ _2F_1\left(1,a;1+a;\frac{1}{1-z}\right) &= \frac{\Gamma(1+a)}{\Gamma(1)\Gamma(a)}\frac{1}{2\pi i} \int_{-i\infty}^{i\infty} \frac{\Gamma(1+s)\Gamma(a+s)\Gamma(-s)}{\Gamma(1+a+s)}(z-1)^{-s}ds \cr &=\frac{a}{2\pi i}\int_{-i\infty}^{+i\infty}\frac{\Gamma(1+s)\Gamma(-s)}{s+a}(z-1)^{-s}ds. \end{align*} In the second relation, change $s$ to $-s$, we get (keeping track of all the minus signs) $$ \ _2F_1\left(1,a;1+a;\frac{1}{1-z}\right) = \frac{a}{2\pi i}\int_{-i\infty}^{+i\infty}\frac{\Gamma(1-s)\Gamma(s)}{a-s}(z-1)^{s}ds. $$ Finally using the relation $$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin \pi s} \quad \mbox{and}\quad \Gamma(-s)\Gamma(1+s)=-\frac{\pi}{\sin \pi(-s)} = -\frac{\pi}{\sin \pi s},$$ we get $$\ _2F_1(1,-a;1-a;{1-z}) = \ _2F_1\left(1,a;1+a;\frac{1}{1-z}\right)$$ and similar the equivalence for the rest two (or just change $a$ to $-a$).

  • $\begingroup$ It seems that your derivation is correct. But I tried to series expand 2F1[1,-a,1-a,1-z]-2F1[1,1,1+a,1/(1-z)], the result is not zero. Do you know why? Thank you so much. $\endgroup$ – Nahc Dec 28 '15 at 19:46
  • $\begingroup$ As I said at the very beginning, the series expansion of the hypergeometric function $\ _2F_1(a,b;c;z)$ converges only for $|z|<1$. You have to use analytical continuation to make sense of it, if $z$ is outside of the unit disk. $\endgroup$ – yhhuang Dec 28 '15 at 20:32

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