In an answer to a question found here @user97357329 implies the following integral $$\int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x)}{1 + x} \, dx,$$ can be found relatively easily.

So far what I have managed to come up with is the following. Since $$\sum_{n = 1}^\infty H^{(2)}_n x^n = \frac{\operatorname{Li}_2 (x)}{1 - x},$$ where $H^{(2)}_n = \sum_{k = 1}^n \frac{1}{k^2}$ denotes the 2nd order generalised harmonic number, replacing $x$ with $-x$ gives $$\sum_{n = 1}^\infty (-1)^n H^{(2)}_n x^n = \frac{\operatorname{Li}_2 (-x)}{1 + x}.$$ So the integral becomes \begin{align} \int_0^1 \frac{\ln (1 - x) \operatorname{Li}_2 (-x)}{1 + x} \, dx &= \sum_{n = 1}^\infty (-1)^n H^{(2)}_n \int_0^1 x^n \ln (1 - x) \, dx\\ &= \sum_{n = 2}^\infty (-1)^{n - 1} H^{(2)}_{n - 1} \int_0^1 x^{n - 1} \ln (1 - x) \, dx\\ &= \sum_{n = 2}^\infty (-1)^n \frac{H^{(2)}_{n - 1} H_n}{n}, \end{align} where the result $\int_0^1 x^{n - 1} \ln (1 - x) \, dx = -\frac{H_n}{n}$ has been used. This gives a difficult non-linear Euler sum.

How can one find the value of the integral without using the value for the Euler sum just found or other difficult non-linear Euler sums (linear ones are fine)?

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    $\begingroup$ You might use that $$\sum_{n=1}^{\infty} x^n H_n H_n^{(2)}$$ $$=\frac{1}{1-x}\biggr(\frac{1}{2}\log(x) \log^2(1-x)+\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)-\zeta(2)\log(1-x)-\zeta(3)\biggr),$$ which is known, and then all calculations keep on flowing naturally. See (Almost) Impossible Integrals, Sums, and Series, page $284$. All the resulting integrals are well-known. $\endgroup$ Jan 30, 2020 at 22:24

2 Answers 2


A solution by Cornel Ioan Valean

Apart from the strategy described in comments, what about if for the last series we combine the following two known identities?

$$\int_0^1 x^{n-1} \log^3(1-x)\textrm{d}x=-\frac{H_n^3+3H_n H_n^{(2)}+2H_n^{(3)}}{n}$$ and $$ \sum_{n=1}^{\infty} x^n(H_n^3-3H_nH_n^{(2)}+2 H_n^{(3)}) = -\frac{\log^3(1-x)}{1-x},$$

which both appear in (Almost) Impossible Integrals, Sums, and Series, pages 2 and 355.

It's easy to see that using the identities above, we have that

$$\sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_n H_n^{(2)}}{n}=-\frac{1}{6} \left(\int_0^1 \frac{\log ^3(1-x)}{1+x} \textrm{d}x+\int_0^1 \frac{\log ^3(1+x)}{x (1+x)} \textrm{d}x\right),$$

where both integrals are straightforward and the desired result follows.

Many thanks to Cornel for this strategy.

UPDATE (July 24, 2022): It is obvious that with the strategy above we can also immediately extract the generating function, $$\sum_{n=1}^{\infty} x^n \frac{H_n H_n^{(2)}}{n}$$ $$=-\zeta(4)+\frac{1}{6}\log(x)\log^3(1-x)-\frac{1}{24}\log^4(1-x)+\frac{1}{2}\log^2(1-x)\operatorname{Li}_2(1-x)$$ $$-\log(1-x)\operatorname{Li}_3(1-x)+\operatorname{Li}_4(1-x)-\operatorname{Li}_4\left(\frac{x}{x-1}\right), \ |x|\le1 \land \ x\neq1.$$

  • $\begingroup$ Yes, that will do it! Following your first suggested approach one ends up having to evaluate 11 different integrals, 2 of which are complex, and, while doable, is by no means trivial. $\endgroup$
    – omegadot
    Jan 31, 2020 at 1:49

Incomplete solution

Start with writing $$\operatorname{Li}_2(-x)=\int_0^1\frac{x\ln y}{1+xy}\ dy$$

$$\Longrightarrow I=\int_0^1\frac{\ln(1-x)\operatorname{Li}_2(-x)}{x(1+x)}\ dx=\int_0^1\ln y\left(\int_0^1\frac{\ln(1-x)}{(1+x)(1+yx)}\ dx\right)\ dy$$

$$=\int_0^1\ln y\left(\frac{\ln^22-\zeta(2)}{2}\cdot\frac{1}{1-y}+\frac{\operatorname{Li}_2\left(\frac{y}{1+y}\right)}{1-y}\right)\ dy$$

$$=-\frac{\ln^22-\zeta(2)}{2}\zeta(2)+\int_0^1\frac{\ln y\operatorname{Li}_2\left(\frac{y}{1+y}\right)}{1-y}dy$$

$$\overset{IBP}{=}\underbrace{\frac54\zeta(4)-\frac12\ln^22\zeta(2)}_{\Large a}-\int_0^1\frac{\operatorname{Li}_2(1-y)\ln(1+y)}{y(1+y)}\ dy$$

$$=a-\underbrace{\int_0^1\frac{\operatorname{Li}_2(1-y)\ln(1+y)}{y}\ dy}_{\large I_1}+\underbrace{\int_0^1\frac{\operatorname{Li}_2(1-y)\ln(1+y)}{1+y}\ dy}_{\large I_2}$$

By integration by parts we have

$$I_1=\int_0^1\frac{\operatorname{Li}_2(-y)\ln y}{1-y}\ dy=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1\frac{x^n\ln y}{1-y}\ dy$$


For $I_2$ use $\operatorname{Li}_2(1-y)=\zeta(2)-\ln y\ln(1-y)-\operatorname{Li}_2(y)$

$$\Longrightarrow I_2=\zeta(2)\int_0^1\frac{\ln(1+y)}{1+y}\ dy-\color{blue}{\int_0^1\frac{\ln y\ln(1-y)\ln(1+y)}{1+y}\ dy}-\int_0^1\frac{\operatorname{Li}_2(y)\ln(1+y)}{1+y}\ dy$$

For the last integral, apply integration by parts

$$\Longrightarrow \int_0^1\frac{\operatorname{Li}_2(y)\ln(1+y)}{1+y}\ dy=\frac12\int_0^1\frac{\ln^2(1+y)\ln(1-y)}{y}\ dy$$

which is a well-known integral. I was not able to calculate the blue integral without using harmonic series, maybe you can take care of it? I hope you find my approach useful.

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    $\begingroup$ For the blue integral we can use $4ab=(a+b)^2-(a-b)^2$ to get: $$4\text{Blue}=\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1+x}dx-\int_0^1 \frac{\ln x \ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx$$ The second integral reduces to something simpler after the substiution $\frac{1-x}{1+x}$ and for the first integral we can use $\frac{1}{1+x}=\frac{1}{1-x^2}-\frac{x}{1-x^2}$ and it reduces to Beta function derivatives. $\endgroup$
    – Zacky
    Jan 31, 2020 at 17:08
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    $\begingroup$ @Zacky very nice . You can post your solution for this integral. Actually I used this algebraic identity but got stuck with the first one you got. $\endgroup$ Jan 31, 2020 at 17:10
  • $\begingroup$ I'll try to post if I find an alternative to derivatives for the Beta function, since I don't really like that. $\endgroup$
    – Zacky
    Jan 31, 2020 at 17:12
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    $\begingroup$ @Zacky we can do without using beta function as have a nice rule here math.stackexchange.com/questions/3402183/… $\endgroup$ Jan 31, 2020 at 17:15

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