Question Number 148887 by mathdanisur last updated on 01/Aug/21

Ω=∫_( 1) ^( ∞)  (((√x) ln x)/(x^2  + 1)) dx = ?

Answered by Kamel last updated on 01/Aug/21

Ω=∫_( 1) ^( ∞)  (((√x) ln x)/(x^2  + 1)) dx =^(t=(1/x)) − ∫_0 ^1 ((Ln(t))/( (√t)(1+t^2 )))dt     =−Σ_(n=0) ^(+∞) (−1)^n ∫_0 ^1 t^(2n−(1/2)) Ln(t)dt=Σ_(n=0) ^(+∞) (((−1)^n )/((2n+(1/2))^2 ))     =Σ_(n=0) ^(+∞) (1/((4n+(1/2))^2 ))−Σ_(n=0) ^(+∞) (1/((4n+(5/2))^2 ))     =(1/(16))(Ψ^((1)) ((1/8))−Ψ^((1)) ((5/8)))

Commented bymathdanisur last updated on 01/Aug/21

Thank you Ser

Answered by Ank0369 last updated on 01/Aug/21

Solution: Ω =^(x=1/y)  ∫_0 ^( 1) −((log_e (y))/( (√y)(1+y^2 ))) dy  =^(y=x^2 )  ∫_0 ^( 1) −4((log_e (x))/(1+x^4 ))dx   −(Ω/4) = Σ_(n=0) ^(n=∞) (−1)^n ∫_0 ^( 1) x^(4n) ∙log_e (x)dx ⇒ Σ_(n=0) ^(n=∞) (((−1)^(n+1) )/((4n+1)^2 ))   −4Ω = Σ_(n=0) ^(n=∞) (((−1)^(n+1) )/((n+(1/4))^2 )) ⇒ (1/4)[ζ(2,(5/8))−ζ(2,(1/8))]  Ω = ∫_1 ^∞ (((√x)∙log_e (x))/(x^2 +1))dx = (1/(16))[ψ^((1)) ((1/8))−ψ^((1)) ((5/8))]