Friday, 25 January 2019

real analysis - Closed-form of sumin=1nftyfrac(1)n+1nPsi3(n+1)=int10fracln(1+x)ln3x1x,dx



Does the following series or integral have a closed-form




n=1(1)n+1nΨ3(n+1)=10ln(1+x)ln3x1xdx





where Ψ3(x) is the polygamma function of order 3.






Here is my attempt. Using equation (11) from Mathworld Wolfram:
Ψn(z)=(1)n+1n!(ζ(n+1)H(n+1)z1)
I got
Ψ3(n+1)=6(ζ(4)H(4)n)
then
n=1(1)n+1nΨ3(n+1)=6n=1(1)n+1n(ζ(4)H(4)n)=6ζ(4)n=1(1)n+1n6n=1(1)n+1H(4)nn=π415ln26n=1(1)n+1H(4)nn
From the answers of this OP, the integral representation of the latter Euler sum is
n=1(1)n+1H(4)nn=1010101010dx1dx2dx3dx4dx5(1x1)(1+x1x2x3x4x5)
or another simpler form
n=1(1)n+1H(4)nn=10Li4(x)x(1+x)dx=10Li4(x)xdx+10Li4(x)1+xdx=Li5(1)10Li4(x)1xdx
I don't know how to continue it, I am stuck. Could anyone here please help me to find the closed-form of the series preferably with elementary ways? Any help would be greatly appreciated. Thank you.







Edit :



Using the integral representation of polygamma function
Ψm(z)=(1)m10xz11xlnmxdx
then we have
n=1(1)n+1nΨ3(n+1)=n=1(1)n+1n10xn1xln3xdx=10n=1(1)n+1xnnln3x1xdx=10ln(1+x)ln3x1xdx
I am looking for an approach to evaluate the above integral without using residue method or double summation.


Answer



Edited: I have changed the approach as I realised that the use of summation is quite redundant (since the resulting sums have to be converted back to integrals). I feel that this new method is slightly cleaner and more systematic.




We can break up the integral into
10ln3xln(1+x)1xdx=10ln3xln(1x)1xdx10(1+x)ln3xln(1x2)(1+x)(1x)dx=10ln3xln(1x)1xdx10ln3xln(1x2)1x2dx10xln3xln(1x2)1x2dx=151610ln3xln(1x)1xdx11610x1/2ln3xln(1x)1xdx=15164βa3b(1,0+)1164βa3b(0.5,0+)
After differentiating and expanding at b=0 (with the help of Mathematica),
4βa3b(a,0+)=[Γ(a)Γ(a+b)(1b+O(1))((ψ4(a)2+(γ+ψ0(a))ψ3(a)+3ψ1(a)ψ2(a))b+O(b2))]b=0=12ψ4(a)+(γ+ψ0(a))ψ3(a)+3ψ1(a)ψ2(a)
Therefore,
10ln3xln(1+x)1xdx=1532ψ4(1)+4516ψ1(1)ψ2(1)+132ψ4(0.5)+18ψ3(0.5)ln2316ψ1(0.5)ψ2(0.5)=12ζ(5)+3π28ζ(3)+π48ln2
The relation between ψm(1), ψm(0.5) and ζ(m+1) is established easily using the series representation of the polygamma function.


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