Difference between revisions of "2F1(1/2,1;3/2;-z^2)=arctan(z)/z"
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Revision as of 21:20, 26 June 2016
Theorem
The following formula holds: $${}_2F_1 \left( \dfrac{1}{2}, 1; \dfrac{3}{2} ; -z^2 \right)=\dfrac{\arctan(z)}{z},$$ where ${}_2F_1$ denotes the hypergeometric 2F1 and $\arctan$ denotes the inverse tangent.
Proof
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous): 15.1.5
