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I find out that $\ln (1+z) = z _2F_1(1,1,2,-z)$ and that $\ln (1-z) = -z _2F_1(1,1,2,z)$, but what is $\ln\Big( \dfrac{1+z}{1-z} \Big)$? Is there a possibility to add two $_2F_1$?

I mean what can I do with this $\ln\Big( \dfrac{1+z}{1-z} \Big) = z(_2F_1(1,1,2,-z) + _2F_1(1,1,2,z))$ ?

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$$ \log\left(\frac{1+z}{1-z}\right) = 2 z \;{}_2F_1\left(\frac{1}{2},1;\frac{3}{2};z^2\right) $$

GEdgar
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  • Thank you. So could you tell me which the proper forum is for questions like that? And how did you get that formula? –  Nov 12 '17 at 13:08
  • And how did you get yours? Hypergeometric functions are just funny power series expansions with the ratios of terms given by some rational functions of $k$ (with real roots and poles, unless you don't mind complex parameters) times the argument. So, you should just make sure that the non-zero coefficients form an arithmetic progressions and the ratios of the consecutive coefficients are rational functions of $k$. In your cases the coefficients themselves are rational functions of $k$, so the only headache is to not forget to cancel the"historical" $(k+1)$ when listing the parameters... – fedja Nov 13 '17 at 00:51