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)?

  • 1
    $\begingroup$ 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. $\endgroup$ – user97357329 Jan 30 '20 at 22:24

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.

  • $\begingroup$ 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 '20 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.

  • 1
    $\begingroup$ 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 '20 at 17:08
  • 1
    $\begingroup$ @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. $\endgroup$ – Ali Shadhar Jan 31 '20 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 '20 at 17:12
  • 1
    $\begingroup$ @Zacky we can do without using beta function as have a nice rule here math.stackexchange.com/questions/3402183/… $\endgroup$ – Ali Shadhar Jan 31 '20 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.