Does the following series or integral have a closed-form
∞∑n=1(−1)n+1nΨ3(n+1)=−∫10ln(1+x)ln3x1−xdx
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)z−1)
I got
Ψ3(n+1)=6(ζ(4)−H(4)n)
then
∞∑n=1(−1)n+1nΨ3(n+1)=6∞∑n=1(−1)n+1n(ζ(4)−H(4)n)=6ζ(4)∞∑n=1(−1)n+1n−6∞∑n=1(−1)n+1H(4)nn=π415ln2−6∞∑n=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=∫10∫10∫10∫10∫10dx1dx2dx3dx4dx5(1−x1)(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)1−xdx
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)m∫10xz−11−xlnmxdx
then we have
∞∑n=1(−1)n+1nΨ3(n+1)=−∞∑n=1(−1)n+1n∫10xn1−xln3xdx=−∫10∞∑n=1(−1)n+1xnn⋅ln3x1−xdx=−∫10ln(1+x)ln3x1−xdx
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)1−xdx=∫10ln3xln(1−x)1−xdx−∫10(1+x)ln3xln(1−x2)(1+x)(1−x)dx=∫10ln3xln(1−x)1−xdx−∫10ln3xln(1−x2)1−x2dx−∫10xln3xln(1−x2)1−x2dx=1516∫10ln3xln(1−x)1−xdx−116∫10x−1/2ln3xln(1−x)1−xdx=1516∂4β∂a3∂b(1,0+)−116∂4β∂a3∂b(0.5,0+)
After differentiating and expanding at b=0 (with the help of Mathematica),
∂4β∂a3∂b(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)1−xdx=−1532ψ4(1)+4516ψ1(1)ψ2(1)+132ψ4(0.5)+18ψ3(0.5)ln2−316ψ1(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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