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How to prove $_2F_1\!\left( -\frac34, \frac34; \frac 14; \frac {\sqrt 3} 2 \right) = \sqrt {\frac {\sqrt 3 - 1} 2} - \frac {\Gamma^2(\frac 14)} {\sqrt \pi 2^{3/4} 3^{5/8}}$? (both approximately equal to -1.614289)

I tried using Mathematica and Hypergeometric identities and all attempts of them failed.

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  • $\begingroup$ Equivalent forms: $$ _2F_1\!\left( \frac 12, 1; \frac 14; \frac {\sqrt 3} 2 \right) = \left(4 + 2 \sqrt 3 \right) \left( 1 + \sqrt {\frac {1 + \sqrt 3} \pi} \frac {\Gamma^2\left(\frac14\right)} {2^{7/4} 3^{5/8}} \right) $$ $$ 1 - {_2F_1\!\left( - \frac 12, 1; \frac 14; \frac {\sqrt 3} 2 \right)} = \sqrt {\frac {1 + \sqrt 3} \pi} \cdot \frac {\Gamma^2\left(\frac14\right)} {2^{3/4} 3^{5/8}} $$ $\endgroup$ Commented Sep 19 at 0:48

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