Integral ∫10ln(1−x)ln(1+x)ln2xdx{\large\int}_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx

This problem was posted at I&S a week ago, and no attempts to solve it have been posted there yet. It looks very alluring, so I decided to repost it here:

Prove:
10ln(1x)ln(1+x)ln2xdx=244π2311π472012ln2+2ln2216ln42+π2ln2+π26ln224Li4(12)354ζ(3)+72ζ(3)ln2.

I found a paper where some similar integrals are evaluated: J. A. M. Vermaseren, Harmonic sums, Mellin transforms and Integrals, Int. J. Mod. Phys. A, 14 (1999), 2037-2076, DOI: 10.1142/S0217751X99001032
, but it’s not quite easy to read for me. Maybe it could be of some help for this problem.

Answer

This answer is split into 3 main steps.


Step 1: Expressing the integral as a sum

     10ln(1+x)ln(1x)ln2x dx=j=1(1)jjk=11k10xj+kln2x dx=2j=1(1)jjk=11k(k+j+1)3=2j=1(1)jjk=11(j+1)3k1(j+1)3(k+j+1)1(j+1)2(k+j+1)21(j+1)(k+j+1)3=2j=1(1)jHj+1j(j+1)32j=1(1)j[ζ(2)H(2)j+1]j(j+1)22j=1(1)j[ζ(3)H(3)j+1]j(j+1)


Step 2a: Value of n=1(1)nHnn
n=1(1)nHnn=12ln22π212
See here for the details.


Step 2b: Value of n=1(1)nHnn2
n=1(1)nHnn2=58ζ(3)
See here for the details.


Step 2c: Value of n=1(1)nHnn3
n=1(1)nHnn3=101y[y01x[x0ln(1t)t(t1)dt]dx]dy=2Li4(12)11π4360+112ln42+74ζ(3)ln2π212ln22
Tunk-Fey did a calculation of this type here.


Step 2d: Value of n=1(1)nH(2)nn
n=1H(2)nnxn=x0Li2(t)t(1t)dt=Li3(x)+x0Li2(t)1tdt=Li3(x)Li2(x)ln(1x)x0ln2(1t)tdt=Li3(x)Li2(x)ln(1x)+1x1ln2t1tdt=Li3(x)Li2(x)ln(1x)ln2(1x)lnx+1x12ln(1t)lnttdt=Li3(x)Li2(x)ln(1x)ln2(1x)lnx2Li2(1x)ln(1x)+1x12Li2(t)tdt=Li3(x)Li2(x)ln(1x)ln2(1x)lnx2Li2(1x)ln(1x)+2Li3(1x)2ζ(3)
Therefore
n=1(1)nH(2)nn=Li3(1)Li2(1)ln2ln22ln(1)2Li2(2)ln2+2Li3(2)2ζ(3)=ζ(3)+π212ln2
You can use polylogarithm identities to simplify the last equation. I took the easy way out and used Wolfram Alpha. Note that contour integration is a slightly more efficient method to solve this sum, however this method is required if I want to solve n=1(1)nH(2)nn2 as well.


Step 2e: Value of n=1(1)nH(3)nn

n=1H(3)nnxn=x0Li3(t)t(1t)dt=Li4(x)+x0Li3(t)1tdt=Li4(x)Li3(x)ln(1x)x0ln(1t)Li2(t)tdt=Li4(x)Li3(x)ln(1x)12Li22(x)
Therefore
n=1(1)nH(3)nn=Li4(1)Li3(1)ln212Li22(1)=19π41440+34ζ(3)ln2


Step 2f: Value of n=1(1)nH(2)nn2

This part is rather similar to Tunk-Fey’s answer, so he certainly deserves credit.
     n=1H(2)nn2xn=Li4(x)2ζ(3)lnx+12Li22(x)+ln2(1x)lnxxdx+2Li3(1x)Li2(1x)ln(1x)xdx
The blue integral is
     ln2(1x)lnxxdx=12ln2xln2(1x)ln2xln(1x)1xdx=12ln2xln2(1x)+n=1Hnxnln2x dx=12ln2xln2(1x)+n=1Hn2nxn+1n+1=12ln2xln2(1x)+ln2xn=1Hnxn+1n+12lnxn=1Hnxn+1(n+1)2+2n=1Hnxn+1(n+1)3=2lnxLi3(x)2Li4(x)2lnxn=1Hnn2xn+2n=1Hnn3xn
The orange integral is
      2Li3(1x)Li2(1x)ln(1x)xdx=2Li3(1x)lnx2Li2(1x)lnxln(1x)+2ln(1x)ln2x1xdx=2Li3(1x)lnx2Li2(1x)lnxln(1x)ln2xln2(1x)4lnxLi3(x)+4Li4(x)+4lnxn=1Hnn2xn4n=1Hnn3xn
So
     n=1H(2)nn2xn=3Li4(x)+2Li3(1x)lnx2Li3(x)lnx2ζ(3)lnx+12Li22(x)2Li2(1x)lnxln(1x)ln2xln2(1x)+2lnxn=1Hnn2xn2n=1Hnn3xn+C
Therefore
     n=1(1)nH(2)nn2=3Li4(1)+2Li3(2)ln(1)2Li3(1)ln(1)2ζ(3)ln(1)+12Li22(1)2Li2(2)ln(1)ln(2)ln2(1)ln22+2ln(1)n=1(1)nHnn22n=1(1)nHnn3=17π44804Li4(12)16ln4272ζ(3)ln2+π26ln22
The grey terms miraculously cancel.


Step 3a: Evaluating 2j=1(1)jHj+1j(j+1)3
     2j=1(1)jHj+1j(j+1)3=2j=1(1)jHj+1j(1)jHj+1(j+1)3(1)jHj+1(j+1)2(1)jHj+1j+1=2j=1(1)jHjj+2j=1(1)jj(j+1)+2j=1(1)jHjj3+2+2j=1(1)jHjj2+2+2j=1(1)jHjj3+2=4Li4(12)11π4180+16ln42+72ζ(3)ln254ζ(3)π26ln22π23+2ln224ln2+8


Step 3b: Evaluating \displaystyle -\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}
\begin{align}
-\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}
&=-\frac{\pi^2}{3}\sum^\infty_{j=1}\frac{(-1)^j}{j}-\frac{(-1)^j}{(j+1)^2}-\frac{(-1)^j}{j+1}\\
&=\frac{\pi^2}{3}\ln{2}+\frac{\pi^4}{36}-\frac{\pi^2}{3}+\frac{\pi^2}{3}\ln{2}-\frac{\pi^2}{3}\\
&=\frac{\pi^4}{36}+\frac{2\pi^2}{3}\ln{2}-\frac{2\pi^2}{3}
\end{align}


Step 3c: Evaluating \displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j(j+1)^2}
\begin{align}
& \ \ \ \ \ 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j(j+1)^2}\\
&=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(2)}}{j}-\frac{(-1)^jH_{j+1}^{(2)}}{(j+1)^2}-\frac{(-1)^jH_{j+1}^{(2)}}{j+1}\\
&=4\sum^\infty_{j=1}\frac{(-1)^jH_{j}^{(2)}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^2}+2\sum^\infty_{j=1}\frac{(-1)^jH_j^{(2)}}{j^2}+2+2\\
&=-8{\rm Li}_4\left(\frac{1}{2}\right)+\frac{17\pi^4}{240}-\frac{1}{3}\ln^4{2}-7\zeta(3)\ln{2}-4\zeta(3)+\frac{\pi^2}{3}\ln^2{2}+\frac{\pi^2}{3}\ln{2}-\frac{\pi^2}{6}-4\ln{2}+8\\
\end{align}


Step 3d: Evaluating \displaystyle -2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)}
\begin{align}
-2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)}
&=-2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j}+2\zeta(3)\sum^\infty_{j=1}\frac{(-1)^j}{j+1}\\
&=4\zeta(3)\ln{2}-2\zeta(3)\\
\end{align}


Step 3e: Evaluating \displaystyle 2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j(j+1)}
\begin{align}
2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j(j+1)}
&=2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j}-2\sum^\infty_{j=1}\frac{(-1)^jH_{j+1}^{(3)}}{j+1}\\
&=4\sum^\infty_{j=1}\frac{(-1)^jH_{j}^{(3)}}{j}+2\sum^\infty_{j=1}\frac{(-1)^j}{j(j+1)^3}+2\\
&=-\frac{19\pi^4}{360}+3\zeta(3)\ln{2}-\frac{3}{2}\zeta(3)-\frac{\pi^2}{6}-4\ln{2}+8
\end{align}


Step 4: Obtaining the final result

Summing the results from steps 3a, 3b, 3c, 3d and 3e gives
\int^1_0\ln(1+x)\ln(1-x)\ln^2{x} \ {\rm d}x=24-\frac{4\pi^2}3-\frac{11\pi^4}{720}-12\ln2\\+2\ln^22-\frac16\ln^42+\pi ^2\ln2+\frac{\pi^2}6\ln^22-4\operatorname{Li}_4\!\left(\tfrac12\right)-\frac{35}4\zeta(3)+\frac72\zeta(3)\ln2.
hence completing the proof.

Attribution
Source : Link , Question Author : Oksana Gimmel , Answer Author : Community

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