Question Number 114635 by mathmax by abdo last updated on 20/Sep/20

find ∫_0 ^1  x^4 ln(x)ln(1−x^2 )dx

Commented bymathdave last updated on 20/Sep/20

Commented bymathdave last updated on 20/Sep/20

Commented byTawa11 last updated on 06/Sep/21

great sir

Answered by maths mind last updated on 20/Sep/20

ln(1−x^2 )=−Σ_(k≥0) (x^(2k+2) /(k+1))  =∫_0 ^1 Σ_(k≥0) (x^(2k+6) /((k+1)))ln(x)dx  =−Σ_(k≥0) (1/(k+1))∫_0 ^1 x^(2k+6) ln(x)dx  =Σ_(k≥0) (1/((k+1)(2k+7)^2 ))  Σ((1/(25(k+1)))−(2/(5(2k+7)^2 ))+(a/((2k+7))))  =((((2k+7)^2 −10(k+1)+25a(k+1)(2k+7))/(25(k+1)(2k+7)^2 )))  50a=−4⇒  =(1/(25))Σ_(k≥0) ((1/(k+1))−((10)/((2k+7)^2 ))−(2/((2k+7))))  =(1/(25))Σ_(k≥0) ((1/(k+1))−(2/(2k+7)))−2Σ_(k≥0) (1/((2k+7)^2 ))  =(1/(25))Σ(((2k+7−2k−2)/((2k+7)(k+1))))−(2/(25))Σ_(k≥0) (1/((2k+7)^2 ))  =(1/5)Σ_(k≥0) (1/(2(k+(7/2))(k+1)))−(2/(25))Σ_(k≥0) (1/((2(k+3)+1)^2 ))  =(1/(15))Σ_(k≥0) (((7/2)−1)/((k+(7/2))(k+1)))−(2/(25))Σ_(k≥3) (1/((2k+1)^2 ))  =(1/(15))(Ψ((7/2))−Ψ(1))−(2/(25))((Σ_(k≥0) (1/((2k+1)^2 )))−1−(1/9)−(1/(25)))  Ψ((7/2))=Ψ((1/2))+(2/5)+(2/3)+2=Ψ((1/2))+((46)/(15))  Σ_(k≥0) ((1/(2k+1)))^2 =(3/4)ζ(2)=(π^2 /8)  Ψ((1/2))=−γ−2ln(2)  we get  (1/(15))(−γ−2ln(2)+((46)/(15))−(−γ))−(2/(25))((π^2 /(48))−((259)/(225)))

Commented byTawa11 last updated on 06/Sep/21

great sur