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I am trying to evaluate $$\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(x\right)}{1+x}\:dx$$ But I am not sure what to do since integration by parts is not possible here.

I tried using a trilogarithm functional equation and got the following integrals. $$-\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(1-x\right)}{1+x}\:dx-\underbrace{\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(\frac{x-1}{x}\right)}{1+x}\:dx}_{J}+\zeta (3)\int _0^1\frac{\ln \left(1-x\right)}{1+x}\:dx$$ $$+\frac{1}{6}\int _0^1\frac{\ln ^3\left(x\right)\ln \left(1-x\right)}{1+x}\:dx+\zeta (2)\int _0^1\frac{\ln \left(x\right)\ln \left(1-x\right)}{1+x}\:dx-\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\ln ^2\left(1-x\right)}{1+x}\:dx$$ All the integrals here are actually manageable except $J$.

Any idea on how to evaluate the main integral differently or the integral $J$? Thank you.

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    Just for answer checking: $$I=2 \text{Li}_5\left(\frac{1}{2}\right)+\frac{\pi ^2 \zeta (3)}{3}-\frac{85 \zeta (5)}{16}-\frac{3}{8} \zeta (3) \log ^2(2)-\frac{1}{60} \log ^5(2)+\frac{1}{36} \pi ^2 \log ^3(2)+\frac{1}{360} \pi ^4 \log (2)$$ – Infiniticism Dec 19 '20 at 04:49
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    And $$J=-2 \text{Li}_5\left(\frac{1}{2}\right)-\frac{3 \pi ^2 \zeta (3)}{32}+\frac{311 \zeta (5)}{64}+\frac{7}{16} \zeta (3) \log ^2(2)+\frac{\log ^5(2)}{60}-\frac{11}{72} \pi ^2 \log ^3(2)-\frac{49 \pi ^4 \log (2)}{1440}$$ – Infiniticism Dec 19 '20 at 04:54
  • @Iridescent Thanks for posting these, seems they are quite tough, btw would you please tell me the closed form for $\sum _{n=1}^{\infty }\frac{H_n^3}{\left(2n+1\right)^2}$? –  Jan 11 '21 at 07:52
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    Sure $$S=48 \text{Li}_5\left(\frac{1}{2}\right)+\frac{7 \pi ^2 \zeta (3)}{24}-\frac{1271 \zeta (5)}{32}-\frac{2}{5} \log ^5(2)-\frac{1}{3} \pi ^2 \log ^3(2)+\frac{61}{240} \pi ^4 \log (2)$$ – Infiniticism Jan 11 '21 at 10:50
  • I was able to boil the integral $I$ down to the evaluation of $\int_{0}^{1}\mathrm{d}x,\frac{\ln^{2}{\left(x\right)}\operatorname{Li}_{2}{\left(x\right)}}{1+x}$. Not sure if that helps much, but I feel like that is progress of a sort. – David H Jan 16 '21 at 00:41

1 Answers1

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Not a complete answer,

$$I=\int _0^1\frac{\log^2 \left(x\right)\operatorname{Li}_2\left(x\right)}{1+x}\:dx$$

Using the Cauchy product,

$$\log(1-x) \, \text{Li}_3(x) = \sum_{n=1}^\infty x^n \left( 4 \frac{1}{n^4} - 2 \frac{H_n}{n^3} - \frac{H_n^{(2)}}{n^2} - \frac{H_n^{(3)}}{n} \right)$$ Using,

$$ \boxed{\operatorname{Li}_4(x)=\sum_{k=1}^\infty\frac{x^k}{k^4}} $$

$$\log(1-x) \, \text{Li}_3(x) = 4\operatorname{Li}_4(x)-\sum_{n=1}^\infty x^n \left( 2 \frac{H_n}{n^3} + \frac{H_n^{(2)}}{n^2} + \frac{H_n^{(3)}}{n} \right)$$

$$I=\int _0^1 \frac{1}{1+x}\left[\color{red}{4\operatorname{Li}_4(x)}-\sum_{n=1}^\infty x^n \left( \color{blue}{2 \frac{H_n}{n^3}} + \color{green}{\frac{H_n^{(2)}}{n^2}} + \color{brown}{\frac{H_n^{(3)}}{n}} \right)\right]\,dx$$

$$I=\color{red}{4\int_0^1\frac{\operatorname{Li}_4(x)}{1+x}\,dx}-\left[\sum_{n=1}^\infty \left( \color{blue}{2 \frac{H_n}{n^3}} + \color{green}{\frac{H_n^{(2)}}{n^2}} + \color{brown}{\frac{H_n^{(3)}}{n}} \right)\right]\int _0^1 \frac{x^n}{1+x}\,dx$$

We have,

$$\color{red}{4\int_0^1\frac{\operatorname{Li}_{4}(x)}{1+x}\mathrm{d}x=4\ln(2)\zeta(4)+3\zeta(2)\zeta(3)-\frac{59}{8}\zeta(5)}$$

We have,

$$\int_0^1 \frac{x^n}{1+x} {\rm d} x= \frac{1}{2} \left[ \psi \left( \frac{n}{2} + 1\right) - \psi \left( \frac{n}{2} + \frac{1}{2} \right) \right]$$

Or we have the alternate,

$$\int_0^1 \frac{x^n}{1+x} \, dx = \frac{1}{2} \left( H_{\frac{n}{2}} - H_{\frac{n-1}{2}} \right)$$

Now comes the challenging part of computing,

$$S_1=\color{blue}{\sum_{n=1}^\infty\frac{H_n}{n^3}\psi \left( \frac{n}{2} + 1\right)}$$ $$S_2=\color{blue}{\sum_{n=1}^\infty\frac{H_n}{n^3}\psi \left( \frac{n}{2} + \frac{1}{2} \right) }$$ $$S_3=\color{green}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}\psi \left( \frac{n}{2} + 1\right)}$$ $$S_4=\color{green}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^2}\psi \left( \frac{n}{2} + \frac{1}{2} \right) }$$ $$S_5=\color{brown}{\sum_{n=1}^\infty\frac{H_n^{(3)}}{n}\psi \left( \frac{n}{2} + 1\right)}$$ $$S_6=\color{brown}{\sum_{n=1}^\infty\frac{H_n^{(3)}}{n}\psi \left( \frac{n}{2} + \frac{1}{2} \right) }$$

Reference :

$$\psi\left(n+\frac{1}{2}\right)=2H_{2n}-H_n-\gamma-2\log(2)$$

$$H_n = ψ(n + 1) + γ$$

We do a nice form for the alternate,

$$I=\int _0^1\frac{\log \left(1-x\right)\operatorname{Li}_3\left(x\right)}{x}\:dx$$

$$\frac{\log(1-x) \, \text{Li}_3(x)}{x} = \sum_{n=1}^\infty x^{n-1} \left( 4 \frac{1}{n^4} - 2 \frac{H_n}{n^3} - \frac{H_n^{(2)}}{n^2} - \frac{H_n^{(3)}}{n} \right)$$

$$\int_0^1 \frac{\log(1-x) \, \text{Li}_3(x)}{x}\,dx = \sum_{n=1}^\infty \left( 4 \frac{1}{n^4} - 2 \frac{H_n}{n^3} - \frac{H_n^{(2)}}{n^2} - \frac{H_n^{(3)}}{n} \right)\int_0^1x^{n-1} \,dx$$

$$\int_0^1 \frac{\log(1-x) \, \text{Li}_3(x)}{x}\,dx = \sum_{n=1}^\infty \left( 4 \frac{1}{n^5} - 2 \frac{H_n}{n^4} - \frac{H_n^{(2)}}{n^3} - \frac{H_n^{(3)}}{n^2} \right)$$

$$\int_0^1 \frac{\log(1-x) \, \text{Li}_3(x)}{x}\,dx = 4\zeta(5)-2\color{green}{\sum_{n=1}^\infty \frac{H_n}{n^4}} - \color{red}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}} - \color{blue}{\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}} $$

From here,

$$\color{red}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)}$$

$$\color{blue}{\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}{2}\zeta(5)-2\zeta(2)\zeta(3)}$$

From here,

$$\color{green}{\sum_{n=1}^\infty \frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)}$$

$$\boxed{\int _0^1\frac{\log \left(1-x\right)\operatorname{Li}_3\left(x\right)}{x}\:dx=\zeta(2)\zeta(3)-3\zeta(5)}$$

Reference :

$$\color{red}{\int_0^1 \frac{\log(x)\operatorname{Li}_2(x)}{1-ax}\,dx=\frac{(\operatorname{Li}_2(a))^2}{2a}-2\zeta(2)\frac{\operatorname{Li}_2(a)}{a}+\frac{3\operatorname{Li}_4(a)}{a}}$$

From David H's comment,

Here's an idea,

$$\int_0^1 \frac{\log^2(x)}{1 + x}\mathrm{Li}_2(x) \, dx = \frac32\zeta(2) \zeta(3) - \int_0^1 \frac{\log^2(x)}{1 + x} \sum_{n=1}^\infty \frac{1 - x^n}{n^2} dx$$

$${\because\int_{0}^{1}\frac{(\log(x))^nx^{p-1}}{1+x^q}dx=\frac{1}{q^{n+1}}\beta^{(n)}\left(\frac{p}{q}\right)\implies\int_{0}^{1}\frac{(\log(x))^2}{1+x}dx=\beta^{(2)}\left(1\right)=\frac32\zeta(3)}$$

Amrut Ayan
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