I am trying to evaluate

$$I=\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\ln(1+x)}{x}-\ln2\right)\ dx$$

I encountered this integral while I was trying to calculate the integral

$$\int_0^1\int_0^1\int_0^1\int_0^1\frac{1}{(1+x) (1+y) (1+z)(1+w) (1+ x y z w)} \ dx \ dy \ dz \ dw$$

First of all we can not split the integrand due to divergence, so I used $\sum_{n=1}^\infty H_n^{(2)}x^{n}=\frac{\operatorname{Li}_2(x)}{1-x}$

which gives us

$$I=\sum_{n=1}^\infty H_n^{(2)}\int_0^1 \left(x^{2n-1}\ln(1+x)-\ln 2 \ x^{2n}\right)\ dx$$

$$I=\sum_{n=1}^\infty H_n^{(2)}\left(\frac{H_{2n}-H_n}{2n}-\frac{\ln2}{2n+1}\right)$$

and I don't know how to proceed. I also tried Abel's summation but it got even more complicated. any idea? All different methods are appreciated, Thank you.

  • $\begingroup$ May be it will be useful to use $\mathrm{Li}_{2}(x^2)=\sum_{k=1}^{\infty}x^{2k}/k^2$. $\endgroup$ Jan 12, 2020 at 10:04
  • $\begingroup$ @Artem Alexandrov I do not think so. $\endgroup$ Jan 30, 2020 at 3:15

1 Answer 1


The development of a solution to the proposed integral

I asked Cornel for a reduction to known integrals and series and here is a possible way to go

First Step

Note that $$\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\ln(1+x)}{x}-\log(2)\right)\textrm{d}x$$ $$=\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\log((1+x)/2)}{x}+\log(2)\frac{1-x}{x}\right)\textrm{d}x$$ $$=\log(2)\int_0^1 \frac{\operatorname{Li}_2\left(x^2\right)}{x(1+x)} \textrm{d}x-\int _0^1 \left(\int _x^1\frac{\text{Li}_2\left(x^2\right)}{x(1-x^2)(1+y)}\textrm{d}y\right)\textrm{d}x$$ $$=\log(2)\underbrace{\int_0^1 \frac{\operatorname{Li}_2\left(x^2\right)}{x(1+x)} \textrm{d}x}_{\text{Reducible to known integrals and series}}-\underbrace{\int _0^1 \left(\int _0^y\frac{\text{Li}_2\left(x^2\right)}{x(1-x^2)(1+y)}\textrm{d}x\right)\textrm{d}y}_{\displaystyle \mathcal{I}}.$$

Second Step

Since we have that $\displaystyle \sum_{n=1}^{\infty} x^n H_n^{(2)} =\frac{\operatorname{Li}_2(x)}{1-x}$ and $\displaystyle \int_0^1\frac{x^{2n}}{1+x}\textrm{d}x=H_{n}-H_{2n}+\log(2),$ then our integral $\mathcal{I}$ reduces to the harmonic series

$$\mathcal{I}=\frac{1}{2}\sum _{n=1}^{\infty } \frac{H_n^{(2)} }{n}\left(H_n-H_{2 n}+\log (2)\right).$$

Third step

We want to apply Abel's summation to the last series and then we get that $$\mathcal{I}=\zeta(2)-\frac{1}{2}\zeta (3)-\frac{1}{2}\log (2)\zeta (3)-\frac{23}{16}\zeta(4)-2 \log ^2(2)$$ $$+\frac{1}{2} \sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(n+1)^2}+\frac{1}{2} \underbrace{\sum _{n=1}^{\infty } \frac{H_n H_n^{(2)}}{(2 n+1) (2 n+2)}}_{\displaystyle \mathcal{S}}.$$ The first series with $n$ instead of $n+1$ in denominator is calculated here.

Fourth step

We may recall 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 also found and proved in the book (Almost) Impossible Integrals, Sums, and Series.

Thus, using this result where we replace $x$ by $x^2$, throwing a double integral over both sides in the form $\displaystyle \int_0^1 \int_0^y f(x)\textrm{d}x\textrm{d}y$, and then rearranging and calculating the elementary resulting integrals, everything reduces to $$\mathcal{S}=\sum _{n=1}^{\infty } \frac{H_n H_n^{(2)}}{(2 n+1) (2 n+2)}=\frac{5 }{4}\zeta (4)- \log (2)\zeta (3)-\log ^2(2)\zeta (2)$$ $$+\underbrace{\int_0^1 \frac{\operatorname{Li}_3\left(x^2\right)}{1+x} \textrm{d}x}_{\text{Reducible to known integrals and series}}+ \underbrace{\int_0^1 \frac{\operatorname{Li}_3\left(1-x^2\right)}{1+x} \textrm{d}x}_{\text{Reducible to known integrals and series}}$$ $$+\underbrace{ \int_0^1 \frac{\log (x) \log ^2\left(1-x^2\right)}{1+x} \textrm{d}x}_{\text{Reducible to known integrals and series}}.$$

A first short note: The case $\displaystyle \sum_{n=0}^\infty(-1)^n(\overline{H}_n-\log(2))^5$ may be reduced to a similar integral which can be approached in a similar style, except that we'll need to deal with weight $5$ harmonic series, and some of them might be pretty challenging like you may see in this paper On the calculation of two essential harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$.

A second short note: The last integral may also be viewed in the form of a Beta function sum since $$ \int_0^1 \frac{\log (x) \log ^2\left(1-x^2\right)}{1+x} \textrm{d}x=\int_0^1 \frac{\log (x) \log ^2\left(1-x^2\right)}{1-x^2}(1-x) \textrm{d}x$$ $$=\frac{1}{4} \underbrace{\int_0^1 \frac{\log (x) \log ^2(1-x)}{\sqrt{x} (1-x)} \textrm{d}x}_{\text{Beta function}}-\frac{1}{4} \underbrace{\int_0^1 \frac{\log (x) \log ^2(1-x)}{1-x} \textrm{d}x}_{\text{Beta function}},$$

and it's also good to see the second integral works nicely by letting $x\mapsto 1-x$ and using geometric series. Good to know the first integral can be finished very nicely with Cornel's Master Theorem of Series, and thus we avoid tedious calculations with Beta function.

A third short note: A good point for the second integral from the Fourth Step (see the final part). Now, if you start with integration by parts and then use the Dilogarithm reflection formula, $\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x)$, you arrive at a bunch of integrals where you might also find very useful to know that $$ \int_0^1 \frac{\displaystyle \log(1-x)\operatorname{Li}_2\left(\frac{x}{x-1}\right)}{1+x} \textrm{d}x=\frac{29}{16} \zeta (4)+\frac{1}{4}\log ^2(2) \zeta (2) -\frac{1}{8} \log ^4(2),$$ which is presented and calculated in (Almost) Impossible Integrals, Sums, and Series, see page $17$ (also you may see a different approach here). Also good to know this is a very important integral in the extraction process of some harmonic series with skew-harmonic numbers!

So, it's clear now we may extract the value of $\displaystyle \int_0^1 \frac{\displaystyle \log(1-x)\operatorname{Li}_2(x)}{1+x} \textrm{d}x$ by also using Landen's Identity.

Alternatively, if you have a taste for Fourier series, we may easily arrive at the form

$$\int_0^1 \frac{\operatorname{Li}_3\left(1-x^2\right)}{1+x} \textrm{d}x= \frac{1}{4} \int_0^1 \frac{\log (1-x) \log ^2(x)}{x} \textrm{d}x+4\int_0^{\pi/2} x \log ^2\left(\sin(x)\right) \textrm{d}x,$$

where in the calculations I used the integral representation of the Trilogarithm you may found on page $4$ in the mentioned book. The second integral in the right-hand side is easily manageable by Fourier series and it has also been calculated here tough definite integral: $\int_0^\frac{\pi}{2}x\ln^2(\sin x)~dx$.

So, there are at least two elegant ways to deal with

$$\int_0^1 \frac{\operatorname{Li}_3\left(1-x^2\right)}{1+x} \textrm{d}x.$$

The closed-form of a challenging auxiliary integral $$\int_0^1 \frac{\operatorname{Li}_3\left(1-x^2\right)}{1+x} \textrm{d}x=2 \log ^2(2)\zeta (2)-\frac{23 }{8}\zeta (4)+\frac{1}{6}\log ^4(2)+4 \operatorname{Li}_4\left(\frac{1}{2}\right).$$

The closed-form of the series $\mathcal{S}$ $$\sum _{n=1}^{\infty } \frac{H_n H_n^{(2)}}{(2 n+1) (2 n+2)}=\frac{19 }{8}\zeta (4)-\frac{1}{6} \log ^4(2)-4 \operatorname{Li}_4\left(\frac{1}{2}\right).$$

The final closed-form

$$\int_0^1\frac{\operatorname{Li}_2(x^2)}{1-x^2}\left(\frac{\ln(1+x)}{x}-\log(2)\right)\textrm{d}x$$ $$=\frac{1}{6}\log ^4(2)-\frac{7 }{2}\zeta (4)+\frac{7}{2}\log (2)\zeta (3)-\frac{3}{2}\log ^2(2)\zeta (2)+4 \operatorname{Li}_4\left(\frac{1}{2}\right).$$

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    $\begingroup$ Awesome (+1). I knew it its a tough integral. $\endgroup$ Jan 12, 2020 at 15:55
  • $\begingroup$ you used the notation $\mathcal{I}$ for two integrals, you meant that? $\endgroup$ Jan 12, 2020 at 16:01
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    $\begingroup$ @AliShather Thanks! No, it's fixed now. $\endgroup$ Jan 12, 2020 at 16:02
  • $\begingroup$ do you have the final closed form? $\endgroup$ Jan 12, 2020 at 19:20
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    $\begingroup$ Thank you :) I appreciate your hard work. $\endgroup$ Jan 12, 2020 at 19:25

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