8

Using Integer Relation Algorithms, I was able to arrive at the following :

$$12\zeta^2(3)=-40\int_{0}^{1}\frac{\ln s\ln^{4}(1+s)}{1+s}ds+40\int_{0}^{1}\frac{\ln^{2}s\ln^{3}(1+s)}{1+s}ds-44\int_{0}^{1}\frac{\ln^{3}s\ln^{2}(1+s)}{1+s}ds+23\int_{0}^{1}\frac{\ln^{4}s\ln(1+s)}{1+s}ds$$

Is it possible to simplify the Right Hand Side so that we have a nice Integral Representation for $\zeta^2(3)?$

Maybe we can apply some Powerful Substitution.

The Coefficients do have Alternating Signs.

I thought to make a pattern appear by removing the powers of $2$ as : $$5\cdot2^3,\ 10\cdot2^2,\ 22\cdot2^1,\ 23\cdot2^0$$

But this does not appear to be binomial.

Also, I do not expect it to be feasible considering the Coefficients do not exhibit any Binomial Pattern.

FDP
  • 13,647
Miracle Invoker
  • 3,200
  • 1
  • 3
  • 22
  • This is beautiful ! The first integral has a closed form. For the other ??? – Claude Leibovici Oct 19 '23 at 07:56
  • @ClaudeLeibovici Actually I had made a Conjecture based on numerical evaluations, using similar Integrals of the form $J(m, n)$ I was able to arrive at Integrals for $\zeta^2(3)\zeta(2)$, $\zeta^2(3)\zeta(4)$. The one presented here is the simplest case. Also I do not know the Closed Form of the Second Integral, but looking at the first one on Wolfram it seems to be quite complicated. – Miracle Invoker Oct 19 '23 at 08:03
  • 1
    We have, for example $$\int_{0}^{1}\frac{\ln^{2}s\ln^{3}1+s}{1+s}ds = \frac{1024}{23} \Re(\text{Li}_6(\frac{1}{2}+\frac{i}{2}))+\frac{472}{23}\text{Li}_6\left(\frac{1}{2}\right)+12 \text{Li}_4(\frac{1}{2}) \log ^2(2)+24 \text{Li}_5(\frac{1}{2}) \log (2)-\frac{3 \zeta (3)^2}{4}+\frac{7}{2} \zeta (3) \log ^3(2)+\frac{93}{8} \zeta (5) \log (2)-\frac{3221 \pi ^6}{60480}+\frac{341 \log ^6(2)}{1035}-\frac{101}{414} \pi ^2 \log ^4(2)-\frac{343 \pi ^4 \log ^2(2)}{16560}$$ this is in a sense the simplest possible (if only $\text{Li}_n$ is allowed). Similar results exist for other terms on RHS. – pisco Oct 19 '23 at 09:26
  • 1
    @pisco. Would you write an answer ? The problem is nice. Thanks in advance & cheers :-) – Claude Leibovici Oct 19 '23 at 12:57
  • @pisco may I ask how did you derive that result? – Zima Oct 20 '23 at 17:03
  • @Black Emperor: You want a proof of your relation? – FDP Oct 21 '23 at 17:24
  • In low weight, $\displaystyle 3\zeta(5)=-16W_{1,3}+12W_{2,2}-8W_{3,1},W_{m,n}=\int_0^1\frac{\ln^m x\ln^n(1+x)}{1+x}dx$ – FDP Oct 21 '23 at 17:36
  • @FDP Yes! A proof would be nice. Also the $\zeta(5)$ relation is nice, using system of equations I have extracted Zeta Values till Weight 10 from integrals of these form, though proving them would be rather hard. – Miracle Invoker Oct 21 '23 at 18:04
  • This leads to the following conjecture. If $\displaystyle W_{m,n}=\int_0^1 \frac{\ln^m x\ln^n x}{1+x}dx$ , $\displaystyle \zeta(n)^2=\sum_{i,j,i+j=2n-1}\underbrace{\lambda_{i,j}}{\in\mathbb{Q}}W{i,j}$ – FDP Oct 21 '23 at 18:11
  • @FDP Nice! I had made the same conjecture but it actually broke! I wasn't able to extract $\zeta^2(5)$. I don't think it was due to numerical problems as I had no trouble extracting $\zeta(2)\zeta(3)\zeta(5)$ and $\zeta^2(3)\zeta(4)$ – Miracle Invoker Oct 21 '23 at 18:15
  • @Zima Probably very difficult to prove using elementary manipulations alone. But this is a consequence of systematic approach, see my answer https://math.stackexchange.com/questions/4758677 or https://math.stackexchange.com/questions/3763243. – pisco Oct 22 '23 at 13:37
  • I don't see any try from you, and you still got upvotes, it's weird because many other people post their problems and they simply didn't know how to solve and want to be helped and got votes to close. – OnTheWay Oct 22 '23 at 15:12
  • 2
    @OnTheWay I mean there are posts with $100$s of upvotes with just a problem statement so I don't know why you would single out this particular post. Also I think my question is being misinterpreted a bit, I wanted to convert the right side into a nice single integral but it seems impossible. Though thanks for the criticism, I will make try to make better posts from now on. – Miracle Invoker Oct 22 '23 at 16:12
  • Another conjecture: If $\displaystyle W_{m,n}=\int_0^1 \frac{\ln^m (1+x)\ln^n x}{1+x}dx$ and, $\displaystyle K_{n}=\int_0^1 \frac{\ln(1-x)\ln^n x}{1-x}dx$ then is $\displaystyle K_n=\sum_{i,j,i+j=n}\underbrace{\lambda_{i,j}}{\in\mathbb{Q}}W{i,j}$?? ($K_n$ is computable) – FDP Oct 22 '23 at 22:05

1 Answers1

5

For $m,n$, positive integers, \begin{align}W_{m,n}=\int_0^{\frac{1}{2}}\frac{\ln^m(1-x)\ln^n x}{1-x}dx\end{align}

\begin{align} K&=\int_0^1\frac{\ln(1-x)\ln^4 x}{1-x}dx\\ J_1&=\int_0^1\frac{\ln^4 (1+x)\ln x}{1+x}dx\overset{u=\frac{x}{1+x}}=-W_{5,0}+W_{4,1}\\ J_2&=\int_0^1\frac{\ln^3 (1+x)\ln^2 x}{1+x}dx\overset{u=\frac{x}{1+x}}=-W_{5,0}+2W_{4,1}-W_{3,2}\\ J_3&=\int_0^1\frac{\ln^2 (1+x)\ln^3 x}{1+x}dx\overset{u=\frac{x}{1+x}}=-W_{5,0}+3W_{4,1}-3W_{3,2}+W_{2,3}\\ J_4&=\int_0^1\frac{\ln (1+x)\ln^4 x}{1+x}dx\overset{u=\frac{x}{1+x}}=-W_{5,0}+4W_{4,1}-6W_{3,2}+4W_{2,3}-W_{1,4}\\ S&=-40J_1+40J_2-44J_3+23J_4=\boxed{21W_{5,0}-46W_{3,2}+48W_{2,3}-23W_{1,4}}\\ \int_0^{\frac{1}{2}}\frac{\ln^5\left(\frac{x}{1-x}\right)}{1-x}dx&=\boxed{-W_{5,0}+5W_{4,1}-10W_{3,2}+10W_{2,3}-5W_{1,4}+W_{0,5}} \end{align} Since, \begin{align}W_{5,0}&=-\frac{\ln^6 2}{6}\\ W_{4,1}&\overset{\text{IBP}}=-\frac{1}{5}\Big[\ln^5(1-x)\ln x\Big]_0^{\frac{1}{2}}+\frac{1}{5}\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^5(1-x)}{x}dx}_{u=1-x}\\ &=-\frac{\ln^6 2}{5}-\frac{8\pi^6}{315}-\frac{W_{0,5}}{5}\\ W_{3,2}&\overset{\text{IBP}}=-\frac{1}{4}\Big[\ln^4(1-x)\ln^2 x\Big]_0^{\frac{1}{2}}+\frac{1}{2}\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^4(1-x)\ln x}{x}dx}_{u=1-x}\\ &=-\frac{\ln^6 2}{4}+\frac{1}{2}\underbrace{\int_0^1 \frac{\ln(1-u)\ln^4 u}{1-u}du}_{=K}-\frac{1}{2}W_{1,4}\\ \end{align} then, \begin{align}S&=\boxed{8\ln^6 2-23K+48W_{2,3}}\\ \int_0^{\frac{1}{2}}\frac{\ln^5\left(\frac{x}{1-x}\right)}{1-x}dx&=\boxed{10W_{2,3}+\frac{5\ln^6 2}{3}-5K-\frac{8\pi^6}{63}}\\ \frac{S}{48}-\frac{1}{10}\int_0^{\frac{1}{2}}\frac{\ln^5\left(\frac{x}{1-x}\right)}{1-x}dx&=\frac{K}{48}+\frac{4\pi^6}{315}\\ S&=K+\frac{64\pi^6}{105}+\frac{24}{5}\underbrace{\int_0^{\frac{1}{2}}\frac{\ln^5\left(\frac{x}{1-x}\right)}{1-x}dx}_{u=\frac{x}{1-x}}\\ &=\boxed{\frac{2\pi^6}{105}+K} \end{align}

Computation of $K$.

\begin{align}K&\overset{\text{IBP}}=\left[\ln(1-x)\left(\int_0^x\frac{\ln^4 t}{1-t}dt-\int_0^1\frac{\ln^4 t}{1-t}dt\right)\right]_0^1+\\&\int_0^1\int_0^1\left(\frac{x\ln^4(tx)}{(1-x)(1-tx)}- \frac{\ln^4 t}{(1-x)(1-t)}\right)dtdx\\ &=\int_0^1\int_0^1\left(\frac{\ln^4 x+4\ln^3x\ln t+6\ln^2 x\ln^2 t+4\ln x\ln^3 t}{(1-x)(1-t)}-\frac{\ln^4(tx)}{(1-t)(1-tx)}\right)dtdx\\ &\overset{\text{Fubini}}=24\zeta(3)^2+\frac{4\pi^6}{45}+\int_0^1\int_0^1\left(\frac{\ln^4 x}{(1-x)(1-t)}-\frac{\ln^4(tx)}{(1-t)(1-tx)}\right)dtdx\\ &=24\zeta(3)^2+\frac{4\pi^6}{45}+\int_0^1\left(\frac{1}{1-t}\left(\int_0^1 \frac{\ln^4 x}{1-x}dx\right)-\frac{1}{t(1-t)}\left(\int_0^t\frac{\ln^4 u}{1-u}du\right)\right)dt\\ &=24\zeta(3)^2+\frac{4\pi^6}{45}+\int_0^1\left(\frac{1}{1-t}\left(\int_t^1 \frac{\ln^4 x}{1-x}dx\right)-\frac{1}{t}\left(\int_0^t\frac{\ln^4 u}{1-u}du\right)\right)dt\\ &\overset{\text{IBP}}=24\zeta(3)^2+\frac{4\pi^6}{45}-K+\underbrace{\int_0^1\frac{\ln^5 t}{1-t}dt}_{=-\frac{8\pi^6}{63}}\\ K&=\boxed{12\zeta(3)^2-\frac{2\pi^6}{105}} \end{align}

Therefore, \begin{align}\boxed{S=12\zeta(3)^2}\end{align}

Addendum

Bonus.

\begin{align}\boxed{\int_0^{\frac{1}{2}} \frac{\ln^2(1-x)\ln^3 x}{1-x}dx=-\frac{\ln^6 2}{6}-\frac{23\pi^6}{2520}+6\zeta(3)^2}\end{align}

FDP
  • 13,647