View source for 2F1(1/2,1;3/2;z^2)=log((1+z)/(1-z))/(2z)

==Theorem==
The following formula holds:
$${}_2F_1 \left( \dfrac{1}{2}, 1 ; \dfrac{3}{2}; z^2 \right)= \dfrac{1}{2z} \log \left( \dfrac{1+z}{1-z} \right),$$
where ${}_2F_1$ denotes the [[hypergeometric 2F1]] and $\log$ denotes the [[logarithm]].

==Proof==

==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=2F1(1,1;2;z)=-log(1-z)/z|next=2F1(1/2,1;3/2;-z^2)=arctan(z)/z}}: $15.1.4$

[[Category:Theorem]]
[[Category:Unproven]]