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Challenging Integral:

\begin{align} I=\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx&=6\operatorname{Li}_5\left(\frac12\right)+6\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{16}\zeta(5)-\frac{21}{8}\zeta(2)\zeta(3)\\&\quad+\frac{21}8\ln^22\zeta(3)-\ln^32\zeta(2)+\frac15\ln^52 \end{align}

The way I computed this integral is really long as it's based on values of tough alternating Euler sums which themselves long to calculate. I hope we can find other approaches that save us such tedious calculations. Any way, here is my approach:

Using the identity from this solution: $\displaystyle\int_0^1 x^{n-1}\ln^3(1-x)\ dx=-\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$

Multiplying both sides by $\frac{(-1)^{n-1}}{n}$ then summing both sides from $n=1$ to $n=\infty$, gives: \begin{align} I&=\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty-\frac{(-x)^{n}}{n}dx=\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx\\ &=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}+3\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2} \end{align}

We have: \begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}&=-6\operatorname{Li}_5\left(\frac12\right)-6\ln2\operatorname{Li}_4\left(\frac12\right)+\ln^32\zeta(2)-\frac{21}{8}\ln^22\zeta(3)\\&\quad+\frac{27}{16}\zeta(2)\zeta(3)+\frac94\zeta(5)-\frac15\ln^52 \end{align}

\begin{align} \sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}&=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac23\ln^32\zeta(2)+\frac74\ln^22\zeta(3)\\&\quad-\frac{15}{16}\zeta(2)\zeta(3)-\frac{23}8\zeta(5)+\frac2{15}\ln^52 \end{align}

$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$

The proof of the first and second sum can be found here and the third sum can be found here.

By substituting these three sums,we get the closed form of $I$.


Other try is by using the rule: ( see here) $$\int_0^1 \frac{\ln^a(1-x)\ln(1+x)}{x}dx=(-1)^a a! \sum_{n=1}^\infty\frac{H_n^{(a+1)}}{n2^n}$$

We get $\quad\displaystyle I=-6\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}\quad$ and this sum is really hard to crack and I think I made it more complicated this way. All approaches are appreciated.

By the way, the last sum was proposed by Cornel last year on his FB page here but he has not revealed his solution yet.

Thanks.

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  • 1
    $\begingroup$ Do you have any idea on how to compute $\int_0^\frac12 \frac{\operatorname{Li}^2_2(x)}{x}dx$? It's the last piece that I need in order to solve the integral in another way. The case where the upper bound is $1$ instead of $\frac12$ is quite easy:math.stackexchange.com/a/3233489/515527, however this causes me some problems. $\endgroup$ – Zacky Jul 15 '19 at 7:50
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    $\begingroup$ Anyway, it looks like this: $$I=-6\ln 2 \operatorname{Li}_4\left(\frac12\right)+6\operatorname{Li}_2\left(\frac12\right)\operatorname{Li}_3\left(\frac12\right)-6\operatorname{Li}_5\left(\frac12\right)-6\int_0^\frac12 \frac{\operatorname{Li}^2_2(x)}{x}dx$$ Hope I copied it right from my notebook. Of course the values for the dilogarithm and trilogarithm are known, and can be further reduced. $\endgroup$ – Zacky Jul 15 '19 at 7:52
  • $\begingroup$ @Zacky the integral you mentioned was proposed by Cornel before he proposed the sum I linked. They are much related and I think Cornel evaluated the integral first then the sum . He has not revealed the solution of both yet and still standing as a challenge. I tried a lot for that integral but kept going in circles. $\endgroup$ – Ali Shadhar Jul 15 '19 at 8:05
  • $\begingroup$ I see, thanks for the input. $\endgroup$ – Zacky Jul 15 '19 at 9:03
  • $\begingroup$ Your super hard to crack Euler sum can be expressed as $$\sum_{n = 1}^\infty \frac{H^{(4)}_n}{n 2^n} = \operatorname{Li}_5 \left (\frac{1}{2} \right ) + \int_0^{\frac{1}{2}} \frac{\operatorname{Li}_4 (x)}{1 - x} \, dx.$$ Expressing the polylog appearing in the integral as a series, followed by integrating by parts 3 times leads me down the garden path to the following term: $$\int_0^{\frac{1}{2}} x^{n - 1} \ln (1 - x) \, dx.$$ But what is to be done with this term? $\endgroup$ – omegadot Jul 15 '19 at 9:50
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This integral was solved by @Song here along with a related integral using a magical algebraic identity

$$2a^3b = -{b^4 \over 2} -{b^4 + 6a^2b^2\over 2} + 3(a^3b+ab^3) - (a-b)^3b$$

with $a=\ln(1-x)$ and $b=\ln(1+x)$

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