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Question Number 162377 by mnjuly1970 last updated on 29/Dec/21
    prove  that          ψ′′ ((1/4) )= −2π^( 3) − 56 ζ (3 )
provethatψ(14)=2π356ζ(3)
Commented by aleks041103 last updated on 29/Dec/21
ψ′′((1/4))=ψ_2 ((1/4))=(−1)^(2+1) 2!Σ_(k=0) ^∞ (1/((k+(1/4))^(2+1) ))=  =−2Σ_(k=0) ^∞ (1/((k+(1/4))^3 ))=−128Σ_(k=0) ^∞ (1/((4k+1)^3 ))
ψ(14)=ψ2(14)=(1)2+12!k=01(k+14)2+1==2k=01(k+14)3=128k=01(4k+1)3
Commented by amin96 last updated on 29/Dec/21
bravo
bravo
Answered by mindispower last updated on 30/Dec/21
Ψ′′((1/4))=−128Σ_(n≥0) (1/((4n+1)^3 ))  Ψ′′(x)−Ψ′′(1−x)=−2π^3 (1+cot^2 (πx))cot(πx)  Ψ′′((1/4))−Ψ′′((3/4))=−4π^3   ζ(3)=Σ_(n≥0) (1/((4n+1)^3 ))+((1/(4n+3)))^3 +(1/(64(n+1)^3 ))+(1/(8(2n+1)^3 ))  =−((Ψ′′((1/4))+Ψ′′((3/4)))/(64))+(1/8)((1/8)ζ(3)+(7/8)ζ(3))  Ψ′′((3/4))=Ψ′′((1/4))+4π^3   ⇒−((2Ψ′′((1/4))+4π^3 )/(64))+((ζ(3))/8)=ζ(3)  ⇒Ψ′′((1/4))=−56ζ(3)−2π
Ψ(14)=128n01(4n+1)3Ψ(x)Ψ(1x)=2π3(1+cot2(πx))cot(πx)Ψ(14)Ψ(34)=4π3ζ(3)=n01(4n+1)3+(14n+3)3+164(n+1)3+18(2n+1)3=Ψ(14)+Ψ(34)64+18(18ζ(3)+78ζ(3))Ψ(34)=Ψ(14)+4π32Ψ(14)+4π364+ζ(3)8=ζ(3)Ψ(14)=56ζ(3)2π

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