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From results and techniques of Integrating $\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx$ with restricted techniques

i think i can prove that:

$\displaystyle \int_0^1\frac{\ln\left(1-t\right)\ln^3 t}{2-t}dt=-\frac{3}{16}\zeta(5)-\frac{21}{4}\zeta(2)\zeta(3)+\frac{9}{4}\zeta(4)\ln 2-\frac{3}{20}\ln^5 2+18 \text{Li}_5\left(\frac{1}{2}\right)$

I'm wondering if this result is obtainable via beta functions and/or (generalised) harmonic series.

FDP
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  • $$\underbrace{\int_0^1\frac{\ln\left(1-t\right)\ln^3 \left(t\right)}{2-t}dt}_{t=1-x}=\sum _{k=1}^{\infty }\left(-1\right)^{k+1}\int _0^1x^{k-1}\ln \left(x\right)\ln ^3\left(1-x\right):dx$$ Then you can use and differentiate with respect to k $$\int _0^1x^{k-1}\ln ^3\left(1-x\right):dx=-\frac{H_k^3+3H_kH_k^{\left(2\right)}+2H_k^{\left(3\right)}}{k}$$ – Dennis Orton Jul 23 '20 at 22:53
  • Can you show us how you calculated it? – Ali Shadhar Jul 25 '20 at 16:56
  • @Ali Shather: i will show you as soon as possible my computation (no use of harmonic series identities, no use of Fourier series, no use of special functions identities) – FDP Jul 25 '20 at 21:18
  • Very interesting, looking forward to seeing it. – Ali Shadhar Jul 25 '20 at 22:06
  • So my solution below and this solution https://math.stackexchange.com/q/3782149 are not good enough? is it like you not accepting solutions? Plus I have not seen your solution yet. – Ali Shadhar Aug 12 '20 at 08:47
  • Ali Shather: Your solution is ok. Is it important for you to obtain points? Is it important to get them fastly? People don't need me to see your solution is ok. Anyway i don't want to be rude. – FDP Aug 12 '20 at 09:25
  • It's not about points. Let's say I reach 100000000 reputation, then what? it's just rude to not appreciate people's hard work. People invest too much time to get decent solutions here and accepting solution as an answer is an act of appreciation. – Ali Shadhar Aug 12 '20 at 09:45
  • @Ali Shather: lately i have spent hours to terminate a computation. I bet if i am lucky enough i would get +1, who care? I'm glad to have been able to terminate this computation, that is important for me. The system of points encourages people to write short and quick answers to maximize the ratio time investment/number of points. Not sure it's a good thing. Quick and short answers tend to be dirty answers. – FDP Aug 15 '20 at 06:38

1 Answers1

2

$$I=\int_0^1\frac{\ln(1-x)\ln^3x}{2-x}dx=\sum_{n=1}^\infty\frac{1}{2^n}\int_0^1 x^{n-1}\ln(1-x)\ln^3x\ dx$$

$$=\sum_{n=1}^\infty\frac{1}{2^n}\frac{\partial^3}{\partial n^3}\int_0^1 x^{n-1}\ln(1-x)\ dx$$

$$=\sum_{n=1}^\infty\frac{1}{2^n}\frac{\partial^3}{\partial n^3}\left(-\frac{H_n}{n}\right)$$

$$=6\sum_{n=1}^\infty\frac{1}{2^n}\left(\frac{H_n}{n^4}+\frac{H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}+\frac{H_n^{(4)}}{n}-\frac{\zeta(2)}{n^3}-\frac{\zeta(3)}{n^2}-\frac{\zeta(4)}{n}\right)$$

$$\small{=6\left(\sum_{n=1}^\infty\frac{H_n}{n^42^n}+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^32^n}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^22^n}+\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}-\zeta(2)\text{Li}_3\left(\frac12\right)-\zeta(3)\text{Li}_2\left(\frac12\right)-\ln(2)\zeta(4)\right)}$$

The last three series dont need to be calculated individually:

By Cauchy product we have

$$-\ln(1-x)\text{Li}_4(x)=2 \sum_{n=1}^\infty\frac{H_n}{n^4}x^n+\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^3}x^n+\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2}x^n+ \sum_{n=1}^\infty\frac{H_n^{(4)}}{n}x^n-5\text{Li}_5(x)$$

Set $x=1/2$ we have

$$\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^32^n}+\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^22^n}+ \sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}=5\text{Li}_5\left(\frac12\right)+\ln(2)\text{Li}_4\left(\frac12\right)-2\sum_{n=1}^\infty\frac{H_n}{n^42^n}$$

Plugging this back in yields

$$\small{I=6\left(-\sum_{n=1}^\infty\frac{H_n}{n^42^n}+5\text{Li}_5\left(\frac12\right)+\ln(2)\text{Li}_4\left(\frac12\right)-\zeta(2)\text{Li}_3\left(\frac12\right)-\zeta(3)\text{Li}_2\left(\frac12\right)-\ln(2)\zeta(4)\right)}$$

In this link we found

\begin{align} \displaystyle\sum_{n=1}^{\infty}\frac{H_n}{ n^42^n}&=2\operatorname{Li_5}\left( \frac12\right)+\ln2\operatorname{Li_4}\left( \frac12\right)-\frac16\ln^32\zeta(2) +\frac12\ln^22\zeta(3)\\ &\quad-\frac18\ln2\zeta(4)- \frac12\zeta(2)\zeta(3)+\frac1{32}\zeta(5)+\frac1{40}\ln^52 \end{align}

Substituting this result along with using $\text{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$ and $\text{Li}_3(1/2)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$ the closed form follows.

Ali Shadhar
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