Question Number 114777 by mathdave last updated on 21/Sep/20

Answered by maths mind last updated on 21/Sep/20

(√x)=t⇒dx=2tdt  =∫_0 ^(π/2) 2tcl_2 (t)dt  =∫_0 ^(π/2) 2tΣ_(k≥1) ((sin(kt))/k^2 )dt  =2Σ_(k≥1) ∫_0 ^(π/2) ((sin(kt))/k^2 )tdt  ∫_0 ^(π/2) ((sin(kt))/k^2 )tdt=(1/k^2 ){[((−cos(kt))/k)t]_0 ^(π/2) +∫_0 ^(π/2) ((cos(kt))/k)dt}  =(π/2)[((−cos(((kπ)/2)))/k^3 )]+(1/k^4 )(sin(k(π/2)))  sin(((kπ)/2))=0  if k=2m  sin(((kπ)/2))=(−1)^m   if k=2m+1  cos(((kπ)/2))  =1 si k=4m,cos(((kπ)/2))=−1 ifk=4m+2  cos(((kπ)/2))=0 if k=4m+1 or 4m+3  so we get Σ_(k≥1) (π/2).[((−cos(((kπ)/2)))/k^3 )]=Σ_(m≥0) (π/2).[((−cos((4m+2)(π/2)))/((4m+2)^3 ))]Σ_(m≥1) −(π/2)((cos(4m.(π/2)))/((4m)^3 ))    +Σ_(m≥0) ((sin((π/2)(2m+1)))/((2m+1)^4 ))  =Σ_(m≥0) (π/2).(1/(8(2m+1)^3 ))+−(π/2).Σ_(m≥1) (1/(64m^3 ))+Σ_(m≥0) (((−1)^m )/((2m+1)^4 ))  =(π/(16))Σ_(m≥0) (1/((2m+1)^3 ))+Σ_(m≥0) (((−1)^m )/((2m+1)^4 ))  β(s)=Σ_(n≥0) (((−1)^n )/((2n+1)^s ))  Dirichlet Betta function  Σ(((−1)^m )/((2m+1)^4 ))=β(4),Σ(1/((2m+1)^3 ))=(ζ(3)−(1/8)ζ(3))=(7/8)ζ(3)  we get  =(π/(16)).(7/( 8))ζ(3)−(π/(128))ζ(3)+β(4)=((6π)/(128))ζ(3)+β(4)=    ∫_0 ^(π/2) cl_2 ((√t))dt=2∫_0 ^1 tcl_2 (t)dt=2.[((6π)/(128))+β(4)]  =((12π)/(128))ζ(3)+2β(4)=((3π)/(32))ζ(3)+2β(4)

Commented byTawa11 last updated on 06/Sep/21

great sir