How to analytically prove
∑k≥1(−1)kk3Hk=−11π4360+ln42−π2ln2212+2Li4(12)+7ln24ζ(3)
As O.L answer
where
Hk=k∑n≥11n.
Addition
So far I developed the following
∑k≥1Hkk2xk=Li3(x)−Li3(1−x)+log(1−x)Li2(1−x)+12log(x)log2(1−x)+ζ(3)
where Li3(x) is the trilogarithm .
For the derivation see http://www.mathhelpboards.com/f10/interesting-logarithm-integral-5301/
Update
A frined on another site gave the following answer
Answer
∑k ≥ 1(−1)kk3Hk=∞∑k=1(−1)kHk 1k3⏞[12∫10ln2(x)xk−1dx]=12∫10ln2(x)[∞∑k=1Hk(−x)k]dxx=12∫10ln2(x)[−ln(1+x)1+x]dxx=−12∫10ln2(x)ln(1+x)(1+x)xdx=12∫10ln2(x)ln(1+x)1+xdx−12∫10ln2(x)ln(1+x)xdx=16∫103ln2(x)ln(1+x)−3ln(x)ln2(1+x)1+xdx+12∫10ln(x)ln2(1+x)1+xdx+12∫−10Li′2(x)ln2(−x)dx=16∫10ln3(x)1+xdx−16∫10ln3(1+x)1+xdx−16∫10ln3(x1+x)dx1+x−16∫10ln3(1+x)xdx−∫−10Li′3(x)ln(−x)dx=−16∫−10ln3(−x)1−xdx−124ln4(2)−16∫1/20ln3(x)1−xdx+16∫21ln3(x)1−xdx + ∫−10Li′4(x)dx⏟=Li4(−1)=−7720π4
The remaining integrals are evaluated by successive integration by parts. Namely,
∫ln3(±x)1−xdx=−ln(1−x)ln3(±x)−3∫Li′2(x)ln2(±x)dx=−ln(1−x)ln3(±x)−3Li2(x)ln2(±x)+6∫Li′3(x)ln(±x)dx=−ln(1−x)ln3(±x)−3Li2(x)ln2(±x)+6Li3(x)ln(±x)−6∫Li′4(x)dx=−ln(1−x)ln3(±x)−3Li2(x)ln2(±x)+6Li3(x)ln(±x)−6Li4(x)
With (1) and (2):
\bbox[15px,#ffe,border:1px dotted navy]{\ds{ \sum_{k\ \geq\ 1}{\pars{-1}^{k} \over k^{3}}H_{k} = -\,{11 \over 360}\,\pi^{4} - {1 \over 12}\ln^{2}\pars{2}\pi^{2} + {1 \over 12}\,\ln^{4}\pars{2} + 2\,\mrm{Li}_{4}\pars{1 \over 2} + {7 \over 4}\,\ln\pars{2}\zeta\pars{3}}}
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