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Question Number 89593 by M±th+et£s last updated on 19/Apr/20

show that ∫_0 ^(π/2) ln(sec(x)) ln(csc(x)) dx=((π^2 ln^2 (2))/2)−(π^4 /(48))

showthat∫0π2ln(sec(x))ln(csc(x))dx=π2ln2(2)2−π448

Commented by maths mind last updated on 19/Apr/20

=∫_0 ^(π/2) ln(sin(x))ln(cos(x))dx =(1/2)∂_a ∂_b 2∫_0 ^(π/2) sin^a (x)cos^b (x)dx∣_((a,b)=(0,0)) =(1/2)∂_a ∂_b β(((a+1)/2),((b+1)/2))_((a,b)=(0,0)) ∂_x β(x,y)=β(x,y)(ψ(x)−ψ(x+y)) ∂_a ∂_b β(((a+1)/2),((b+1)/2)) =(1/2)∂_a β(((a+1)/2),((b+1)/2))(Ψ(((b+1)/2))−Ψ(((a+b)/2)+1)) =(1/4)β(((a+1)/2),((b+1)/2))(Ψ(((a+1)/2))−Ψ(((a+b)/2)+1))(Ψ(((b+1)/2))−Ψ(((a+b)/2)+1)) −(1/4)Ψ′(((a+b)/2)+1)β(((a+1)/2),((b+1)/2)) (a=b=0) =((β((1/2),(1/2)))/4)((Ψ((1/2))−1)^2 −Ψ′(1)) =(π/8)(4ln^2 (2)−(π^2 /6))=((πln^2 (2))/2)−(π^3 /(48))

=∫0π2ln(sin(x))ln(cos(x))dx=12∂a∂b2∫0π2sina(x)cosb(x)dx∣(a,b)=(0,0)=12∂a∂bβ(a+12,b+12)(a,b)=(0,0)∂xβ(x,y)=β(x,y)(ψ(x)−ψ(x+y))∂a∂bβ(a+12,b+12)=12∂aβ(a+12,b+12)(Ψ(b+12)−Ψ(a+b2+1))=14β(a+12,b+12)(Ψ(a+12)−Ψ(a+b2+1))(Ψ(b+12)−Ψ(a+b2+1))−14Ψ′(a+b2+1)β(a+12,b+12)(a=b=0)=β(12,12)4((Ψ(12)−1)2−Ψ′(1))=π8(4ln2(2)−π26)=πln2(2)2−π348

Commented by M±th+et£s last updated on 19/Apr/20

thank you verry much its great solution

thankyouverrymuchitsgreatsolution

Commented by maths mind last updated on 19/Apr/20

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Commented by M±th+et£s last updated on 19/Apr/20

i am sorry this happened to you and i am verry happy to see you again in the forum

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Commented by M±th+et£s last updated on 20/Apr/20

sir i can put all of your solutions on my questions here in this post if you want that?

siricanputallofyoursolutionsonmyquestionshereinthispostifyouwantthat?

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