...advanced calculus...
evaluation of:
Ω=∫_0 ^( (π/2)) sin(x)log(sin(x))dx
by using the euler beta and gamma function:
β(p,(1/2))=2∫_0 ^(π/2) sin^(2p−1) (x)dx
((dβ(p,(1/2)))/dp) =2∫_0 ^(π/2) 2sin^(2p−1) (x)ln(sin(x))dx
=4∫_0 ^(π/2) sin^(2p−1) (x)ln(sin(x))dx
Ω=(1/4)[((dβ(p,(1/2)))/dp)]_(p=1)
((d(β(p,(1/2))))/dp)=(√π)[((Γ′(p)Γ(p+(1/2))−Γ′(p+(1/2))Γ(p))/(Γ^2 (p+(1/2))))]_(p=1)
Ω=(1/4)((√π)[((Γ^′ (1)Γ((3/2))−Γ′((3/2))Γ(1))/(Γ^2 ((3/2))))])
=((√π)/4)[((−γ(((√π)/2))−((√π)/2)(2−γ−2ln(2)))/(π/4))]
=((√π)/4)∗((√π)/2)(((−γ−2+γ+2ln(2))/(π/4)))
=ln(2)−1
... m.n.july.1970...
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