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a friend posted the following integral

$$I=\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$$

The best we could do is expressing it in terms of Euler sums

$$I=-\frac{\zeta^2(2)}{2}+ \sum_{n\geq 1}\frac{(-1)^{n-1}}{n^2} H_{n}^{(2)}+\sum_{n\geq 1}\frac{(-1)^{n-1}}{n^3}H_{n}$$

I am wondering if the approach I followed made the integral complicated ? What approach would you follow to solve the integral?, can we find a better solution ?

  • $\begingroup$ By the way, why are you interested in Euler sums? $\endgroup$ – Mhenni Benghorbal Aug 14 '13 at 3:19
  • $\begingroup$ @Mhenni Benghorbal, I was working with polylogarithm function and I found an intimate relation with Euler sums . $\endgroup$ – Zaid Alyafeai Aug 14 '13 at 3:25
  • $\begingroup$ @mhenniBenghorbal, sorry I don't get what you are saying ? $\endgroup$ – Zaid Alyafeai Aug 16 '13 at 16:53

The values of the two Euler Sums are

$$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n}{n^{3}} = \frac{11\pi^4}{360}-2\text{Li}_4 \left(\frac{1}{2} \right)-\frac{7}{4}\log(2) \zeta(3)+\frac{\pi^2}{12}\log^2(2)-\frac{1}{12}\log^4(2)$$ $$\displaystyle \sum_{n=1}^\infty (-1)^{n-1} \frac{H_n^{(2)}}{n^{2}} =-\frac{17}{480}\pi^4 +4 \text{Li}_4 \left(\frac{1}{2} \right)+\frac{7}{2}\log(2) \zeta(3)-\frac{\pi^2 \log^2(2)}{6}+\frac{\log^4(2)}{6}$$

Therefore the integral evaluates to

$$\begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \sim 0.290721 \end{align*}$$

Refer to this page for the evaluation of Euler Sums.

  • $\begingroup$ The following relation is useful for your calculations. $\endgroup$ – Mhenni Benghorbal Aug 17 '13 at 19:24

using an identity devoloped by Cornel Ioan Valean and it can be found in his book " Almost impossible integrals, sums, and series": $$\ln(1-x)\ln(1+x)=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)x^{2n} $$ we get: \begin{align} I&=\int_0^1\frac{\ln(1-x)\ln(1+x)\ln x}{x}\ dx=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\int_0^1x^{2n-1}\ln x\ dx\\ &=-\sum_{n=1}^\infty\left(\frac{H_{2n}-H_n}{n}+\frac1{2n^2}\right)\left(-\frac1{(2n)^2}\right)=2\sum_{n=1}^\infty\frac{H_{2n}}{(2n)^3}-\frac14\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)\\ &=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)\\ &=2\operatorname{Li}_4\left(\frac12\right)-\frac12\ln^22\zeta(2)+\frac74\ln2\zeta(3)-\frac{27}{16}\zeta(4)+\frac1{12}\ln^42 \end{align}

where we used the value of the first sum proved here and the common value of the second sum which is $\frac54\zeta(4)$.


Related problems: (I). You can have the following solution

$$ \frac{3\gamma}{4}\,\zeta( 3 )+{\frac {7\pi^4}{360}}+\sum _{m=1}^{\infty }{\frac { \left( -1 \right) ^{m-1}\psi \left( m \right) }{{m}^{3}}}+\sum _{m=1}^{\infty }-{\frac { \left( -1 \right) ^{m-1}\psi' \left( m \right) }{{m}^{2}}}\sim 0.2907212779,$$

which you might be able to simplify it further.

Note: If you use the identity

$$ \frac{\pi^4}{90}=\zeta(4), $$

in the above expression, then you will have the form

$$ \frac{3\gamma}{4}\,\zeta( 3 )+{\frac {7}{4}}\zeta(4)+\sum _{m=1}^{\infty }{\frac { \left( -1 \right) ^{m-1}\psi \left( m \right) }{{m}^{3}}}+\sum _{m=1}^{\infty }-{\frac { \left( -1 \right) ^{m-1}\psi' \left( m \right) }{{m}^{2}}}.$$


I would do the following variable change.

$$x=e^{-t}$$ Then we can represent the integral as follows:

$$I=-\int_{0}^{\infty}t\ln(1+e^{-t})\ln(1-e^{-t})\;dt$$ Now, apply the Taylor expansion of the logarithm:





$$-\frac{\partial^2}{\partial s\partial t}\left[B(s+1,t+1)\;_3 F_2(1,1,s+1;2,s+t+2;-1)\right]_{s=t=0}$$

It may be that the Hypergeometric function is summable. In this case, the differentiation is trivial. (B denotes Euler's beta function.)

  • 3
    $\begingroup$ Your $LaTeX$ got messed up. Try to fix it. $\endgroup$ – Kunnysan Aug 12 '13 at 4:59

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