In my previous post on Euler sums, we evaluated sums containing H_n and H_n^{(2)}. In this post, we’ll derive some further results using the integral \int_0^x \frac{\log^3(1-t)}{t}dt. Our starting point is the following generating function identity: \sum_{n=1}^\infty (H_n)^2 x^n = \frac{\log^2(1-x)+\text{Li}_2(x)}{1-x} ,\quad -1\leq x < 1 This can derived by plugging H_n = \int_0^1\frac{1-t^n}{1-t}dt and interchanging the sum and the integral. We can rewrite the above equation using the fact that \sum_{n=1}^\infty H_n^{(2)} x^n = \frac{\text{Li}_2(x)}{1-x}. \sum_{n=1}^\infty (H_n)^2 x^n = \frac{\log^2(1-x)}{1-x} + \sum_{n=1}^\infty H_n^{(2)} x^n , \quad -1\leq x < 1 \quad \color{blue}{\cdots (1)}

## 1. Evaluation of \int_0^x \frac{\log^n(1-t)}{t}dt

In this section, we will derive a formula for the integral \int_0^x \frac{\log^n(1-t)}{t}dt where n is a positive integer. First, we’ll consider the case when 0\leq x < 1. We have: \begin{aligned} \int_0^x \frac{\log^n (1-t)}{t}dt &= \int_0^{-\log(1-x)} \frac{t^n e^{-t}}{1-e^{-t}}dt \quad (t\mapsto 1-e^{-t}) \\ &= \sum_{j=1}^\infty \int_0^{-\log(1-x)}t^n e^{-jt} dt \\ &= -\sum_{j=1}^\infty \left[e^{-jt}\sum_{i=0}^n \frac{(-1)^{n-i}\log^{n-i}(1-x)}{j^{i+1}}n^{\underline{i}} \right]_0^{-\log(1-x)} \\ &= - \sum_{j=1}^\infty \left( (1-x)^j \sum_{i=0}^n \frac{(-1)^{n-i}\log^{n-i}(1-x)}{j^{i+1}}n^{\underline{i}}-\frac{n!}{j^{n+1}} \right) \\ &= n! \zeta(n+1) + \sum_{i=0}^n (-1)^{n-i+1} n^{\underline{i}}\log^{n-i}(1-x) \text{Li}_{i+1}(1-x) \\ &\quad \color{blue}{\cdots (2)} \end{aligned}

A similar calculation shows that for -1\leq x < 0, we have: \begin{aligned} \int_x^0 \frac{\log^n(1-t)}{t}dt &= -\frac{\log^{n+1}(1-x)}{n+1} - n! \zeta(n+1) + \sum_{i=0}^{n} n^{\underline{i}}\log^{n-i}(1-x)\text{Li}_{i+1}\left(\frac{1}{1-x}\right) \\ &\quad \color{blue}{\cdots (3)} \end{aligned}

## 2. Sums with (H_n)^2

We can divide equation (1) by x and integrate both sides to get some interesting results. \begin{aligned} \sum_{n=1}^\infty \frac{(H_n)^2}{n}x^n &= -\frac{\log^3(1-x)}{3}-\log(1-x)\text{Li}_2(x)+\text{Li}_3(x), \quad -1\leq x < 1\quad \color{blue}{\cdots (4)} \\ \sum_{n=1}^\infty \frac{(H_n)^2}{n^2}x^n &= \text{Li}_4(x) + \frac{\text{Li}_2^2(x)}{2}-\frac{1}{3}\int_0^x \frac{\log^3(1-t)}{t}dt, \quad -1\leq x \leq 1\quad \color{blue}{\cdots (5)} \end{aligned} Equations (4) was obtained with the help of results from section (4) of this post.

One can now plug in x=-1 in equation (5) to get: \boxed{\sum_{n=1}^\infty \frac{(H_n)^2}{n^2}(-1)^{n+1} = \frac{41\pi^4}{1440} + \frac{\pi^2 \log^2(2)}{12}-\frac{\log^4(2)}{12}-\frac{7}{4}\log(3)\zeta(3)-2\text{Li}_2\left(\frac{1}{2}\right)} Of course, equation (3) was used to evaluate \int_{-1}^0 \frac{\log^3(1-t)}{t}dt.