Some series and integrals involving the Riemann zeta function binomial coefficients and the harmonic numbers Volume II

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1、whereitiseasytoshowbyusingIBPthat∫∫Li2(x)Li2(1−x)dx=−dx1−xx2=2Li3(x)−2Li2(x)ln(x)−Li2(1−x)lnx−ln(1−x)lnxand∫∫Li3(x)Li3(1−x)12dx=−dx=−Li2(1−x)−Li3(1−x)lnx.1−xx2Now,allunknowntermshavebeenobtained.PuttingaltogethertoSomeSeriesandIntegralsH4(x),wehave4212ππ1326H

2、4(x)=ζ(3)lnx+lnx−Li3(x)−lnxln(1−x)+Li5(x)1015030605[]InvolvingtheRiemannZeta1121−Li3(x)−Li2(x)lnx−ln(1−x)lnxLi2(1−x)−Li4(x)525311212−Li4(x)lnx+Li3(x)lnx+Li3(x)lnx−Li3(1−x)lnxFunction,BinomialCoefficients55510131(3)1(2)1(3)−Li2(x)lnx−H2(x)+H2(x)+H1(x)lnx155551

3、(2)21213andtheHarmonicNumbers−H1(x)lnx+H3(x)lnx−H2(x)lnx+H1(x)lnx+C.55515Thenextstepisfindingtheconstantofintegration.Settingx=12一些包含黎曼π6ζ函数1,二项式系数1(3)1(2)H4(1)=−Li3(1)+Li5(1)−Li4(1)−H2(1)+H2(1)+C3055552π1933ζ(5)+ζ(2)ζ(3)=−Li3(1)+Li5(1)+Li3(1)+C和调和数的积分与级数30305

4、42ππ3C=+ζ(3)−ζ(3)+3ζ(5).45055ThusVolumeII4212ππ1326H4(x)=ζ(3)lnx+lnx−Li3(x)−lnxln(1−x)+Li5(x)1015030605[]1121−Li3(x)−Li2(x)lnx−ln(1−x)lnxLi2(1−x)−Li4(x)525311212−Li4(x)lnx+Li3(x)lnx+Li3(x)lnx−Li3(1−x)lnx55510131(3)1(2)1(3)−Li2(x)lnx−H2(x)+H2(x)+H1(x)lnx155551

5、(2)21213Everyintegralandseriesharborsa−H1(x)lnx+H3(x)lnx−H2(x)lnx+H1(x)lnx5551542ππ3storyasfascinatingasacomicbook!++ζ(3)−ζ(3)+3ζ(5)450551andsettingx=2()5221ln2π3ζ(3)2πH4=−ln2+ln2−ζ(3)240362124()()ζ(5)π11+−ln2+Li4ln2+2Li5.3272022Finally,weobtain∫Author:DonalF

6、.Connon1ln3(1+x)lnxπ2992π221532dx=ζ(3)+ζ(5)−lnEmail:2+lndconnon@btopenworld.com2−ζ(3)ln2x2165340()()11−12Li4ln2−12Li5221SomeseriesandintegralsinvolvingtheRiemannzetafunction,binomialcoefficientsandtheharmonicnumbersVolumeIVDonalF.Connon13October2007AbstractIn

7、thisseriesofsevenpapers,predominantlybymeansofelementaryanalysis,weestablishanumberofidentitiesrelatedtotheRiemannzetafunction,includingthefollowing:∞(1)xHLnsni()x−Lis()y∑sx=∫dyn=1nx0−y∞(3)Hn2471∑n=−+−ςς(2)log2(3)log2Li4(1/2)log2n=1n286∞nnxk−−1un1nj−−⎛⎞()jnj∫

8、uelogudu=−Γx∑(1)⎜⎟()logxk0kj=0⎝⎠j∞n⎛⎞nH(2)12k25∑∑n+14⎜⎟=+ς(2)(4)ςς(3)−ς(6)nk==1121⎝⎠kk2∞n⎛⎞nH(3)11k21∑∑n+13⎜⎟=+ς(3)ς(6)nk==1122⎝⎠kk2∞nnkH(2)11⎛⎞(1)−k+14229∑∑⎜⎟3=−+ς(3)ςςς(6)(2)(4)31(nk==0