Monday, 3 June 2013

sequences and series - Alternating harmonic sum sumkgeq1frac(1)kk3Hk



How to analytically prove




k1(1)kk3Hk=11π4360+ln42π2ln2212+2Li4(12)+7ln24ζ(3)



As O.L answer



where
Hk=kn11n.



Addition




So far I developed the following



k1Hkk2xk=Li3(x)Li3(1x)+log(1x)Li2(1x)+12log(x)log2(1x)+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[1210ln2(x)xk1dx]=1210ln2(x)[k=1Hk(x)k]dxx=1210ln2(x)[ln(1+x)1+x]dxx=1210ln2(x)ln(1+x)(1+x)xdx=1210ln2(x)ln(1+x)1+xdx1210ln2(x)ln(1+x)xdx=16103ln2(x)ln(1+x)3ln(x)ln2(1+x)1+xdx+1210ln(x)ln2(1+x)1+xdx+1210Li2(x)ln2(x)dx=1610ln3(x)1+xdx1610ln3(1+x)1+xdx1610ln3(x1+x)dx1+x1610ln3(1+x)xdx10Li3(x)ln(x)dx=1610ln3(x)1xdx124ln4(2)161/20ln3(x)1xdx+1621ln3(x)1xdx + 10Li4(x)dx=Li4(1)=7720π4




The remaining integrals are evaluated by successive integration by parts. Namely,





ln3(±x)1xdx=ln(1x)ln3(±x)3Li2(x)ln2(±x)dx=ln(1x)ln3(±x)3Li2(x)ln2(±x)+6Li3(x)ln(±x)dx=ln(1x)ln3(±x)3Li2(x)ln2(±x)+6Li3(x)ln(±x)6Li4(x)dx=ln(1x)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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