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Question Number 122290 by mnjuly1970 last updated on 15/Nov/20
... nice calculus... calculate :: Ω=∫_0 ^( 1) x^2 (ψ(1+x)−ψ(2−x))dx=??? .m.n.1970.
…nicecalculus…calculate::Ω=∫01x2(ψ(1+x)−ψ(2−x))dx=???.m.n.1970.
Commented by Dwaipayan Shikari last updated on 15/Nov/20
2−log(2π)
2−log(2π)
Answered by mnjuly1970 last updated on 16/Nov/20
solution: Φ=∫_0 ^( 1) x^2 ψ(2−x)dx=^(∫_a ^( b) f(x)=∫_a ^( b) f(a+b−x)dx) ∫_0 ^( 1) (1−x)^2 ψ(1+x)dx ∴ Ω=∫_0 ^( 1) x^2 (ψ(1+x))dx−∫_0 ^( 1) (1−x)^2 ψ(1+x)dx =∫_0 ^( 1) (2x−1)ψ(1+x)dx=[(2x−1)lnΓ(x+1)]_0 ^1 −2∫_0 ^( 1) lnΓ(x+1)dx =−2∫_0 ^( 1) ln(x)−2∫_0 ^( 1) ln(Γ(x))dx =2−2((1/2)ln(2π))=2−ln(2π) ✓ ..m.n..
solution:Φ=∫01x2ψ(2−x)dx=∫abf(x)=∫abf(a+b−x)dx∫01(1−x)2ψ(1+x)dx∴Ω=∫01x2(ψ(1+x))dx−∫01(1−x)2ψ(1+x)dx=∫01(2x−1)ψ(1+x)dx=[(2x−1)lnΓ(x+1)]01−2∫01lnΓ(x+1)dx=−2∫01ln(x)−2∫01ln(Γ(x))dx=2−2(12ln(2π))=2−ln(2π)✓..m.n..

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