A work in progress inspired by Ramanujan
This is the case a=16 of 2F1(a,a;a+12;−u)=2aΓ(a+12)√πΓ(a)∫∞0dx(1+2u+coshx)a
14321/4K(k3)∫10dx√1−x6√x5+27ϕ9x6=2F1(16,16;23;−27ϕ9)=355/6ϕ−1
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