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Saturday, May 31, 2025

Entry 121

This is the case a=13 of 2F1(a,a;a+12;u)=2aΓ(a+12)πΓ(a)0dx(1+2u+coshx)a

we have 1481/4K(k3)10dx1x3x2+4x3=2F1(13,13;56;4)=355/6

1481/4K(k3)10dx1x3x2+27x3=2F1(13,13;56;27)=47

(To be continued.) 

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