Is the following identity true? $$ \sum_{n=1}^\infty \frac{H_nx^n}{n^3} = \frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right] + \operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}$$
In this accepted answer, @Tunk Fey proved the above. (See $(4)$). However, I have the following $3$ queries :
- Why can we add the integrals after the substitution $x \mapsto 1-x$ in the following step? I doubt it since $\int f(x) \ \mathrm{d}x \neq \int f(1-x) \ \mathrm{d}x$ in general.
- Why do we omit the constant of integration in the following step? We should add the constant since it will affect the summation.
$$\begin{align} \color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}&=-\int\frac{\ln (1-x)\ln^2 x}{1-x}\ dx\\ &=\int\sum_{n=1}^\infty H_n x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \int x^n\ln^2x\ dx\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\int x^n\ dx\right]\\ &=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\frac {x^{n+1}}{n+1}\right]\\ &=\sum_{n=1}^\infty H_n \left[\frac{x^{n+1}\ln^2x}{n+1}-2\frac{x^{n+1}\ln x}{(n+1)^2}+2\frac{x^{n+1}}{(n+1)^3}\right]\\ &=\ln^2x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{n+1}-2\ln x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^2}+2\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^3}\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^2}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^3}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\sum_{n=1}^\infty\frac{x^{n}}{n^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\sum_{n=1}^\infty\frac{x^{n}}{n^4}\right]\\ &=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\operatorname{Li}_4(x)\right]. \end{align}$$
- Is the identity even true, since putting $x=\dfrac{1}{2}$ gives a numerically different result than the correct result, as pointed out by the user @Super Abound in the comments of that answer.
Please help.
The sum in question has a closed form in terms of polylogarithms. The proof is complicated, and I don't intend to reproduce it as I derived it some 15 years ago, and polylogs are not a primary interest now. You can always differentiate both sides and use polylog IDs in Lewin.
$$\sum_{k=1}^\infty \frac{y^k}{k^3}H_k=\zeta(4)+2 Li_4(y)-Li_4(1-y)+Li_4(-y/(1-y))+\\ \frac{1}{2} \log(1-y) \Big( \zeta(3) – Li_3(y)+Li_3(1-y)+Li_3(-y/(1-y)) \Big) + \\\frac{1}{12}\log^3(1-y)\log(y) -\frac{1}{24}\log^4(1-y)$$