Integrating $\int_0^1 \frac {\log(1-x)\log^2(1+x)}x \mathrm{d}x$

Question

Question: Integrate$$\int\limits_0^1dx\,\frac {\log(1-x)\log^2(1+x)}x=-\frac {\pi^4}{240}$$

I'm curious as to if there is a way to integrate this. I've tried using integration by parts to get$$I=-\frac {\pi^2}6\log^22+2\int\limits_0^1dx\,\frac {\operatorname{Li}_2(x)\log(1+x)}{1+x}$$However, I'm not sure how to continue even further. The polylog in the second integrand seems a bit intimidating and I don't see how the first term even helps.

Answer 1

The integral is hard to tackle directly (without using Euler sums), but there is a nice trick (which is literally the same as posed above).

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Let $$I = \int_0^1 {\frac{{\ln (1 - x){{\ln }^2}(1 + x)}}{x}dx} \qquad J = \int_0^1 {\frac{{{{\ln }^2}(1 - x)\ln (1 + x)}}{x}dx} $$

We have $$\begin{aligned} 3I + 3J + \int_0^1 {\frac{{{{\ln }^3}(1 - x)}}{x}dx} + \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx} &= \int_0^1 {\frac{{{{\ln }^3}(1 - {x^2})}}{x}dx} \\ &= \frac{1}{2}\int_0^1 {\frac{{{{\ln }^3}(1 - u)}}{u}du} \end{aligned}$$ the substitution $x^2 = u$ is used. Hence $$\tag{1}3I + 3J + \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx} = \frac{{{\pi ^4}}}{{30}}$$

On the other hand, $$\begin{aligned}\int_0^1 {\frac{{{{\ln }^3}(1 - x)}}{x}dx} - 3J + 3I - \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx} &= \int_0^1 {\frac{{{{\ln }^3}(\frac{{1 - x}}{{1 + x}})}}{x}dx} \\ &= \int_0^1 {\frac{{2{{\ln }^3}u}}{{(1 - u)(1 + u)}}du} \\ &= \int_0^1 {\frac{{{{\ln }^3}u}}{{1 - u}}du} + \int_0^1 {\frac{{{{\ln }^3}u}}{{1 + u}}du} \end{aligned}$$ the substitution $u=\frac{1-x}{1+x}$ is used. Giving $$\tag{2} - 3J + 3I - \int_0^1 {\frac{{{{\ln }^3}(1 + x)}}{x}dx} = - \frac{{7{\pi ^4}}}{{120}} $$

Adding $(1)$ and $(2)$ together gives $I=-\frac{\pi^4}{240}$.

Answer 2

It's quite common for such integrals to use algebraic identities in order to solve them, see also here.

We can use the first one in the link from above, namely: $$6ab^2=(a+b)^3+(a-b)^3-2a^3\Rightarrow I=\int_0^1 \frac{\ln(1-x) \ln^2(1+x)}{x} dx$$ $$=\frac16\int_0^1 \frac{\ln^3(1-x^2)}{x}dx+\frac16\int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}dx-\frac13\int_0^1 \frac{\ln^3(1-x)}{x}dx$$ For the first one let $1-x^2 =t$, for the second one $\frac{1-x}{1+x}=t$ and for the third one $1-x=t$ to get: $$I=\frac1{12} \int_0^1 \frac{\ln^3 t}{1-t}dt+\frac13\int_0^1 \frac{\ln^3 t}{1-t^2}dt-\frac13\int_0^1 \frac{\ln^3 t}{1-t}dt$$ $$=-\frac14\int_0^1 \frac{\ln^3 t}{1-t}dt+\frac13\int_0^1 \frac{\ln^3 t}{1-t^2}dt$$ $$=-\frac14\sum_{n=1}^\infty \int_0^1 t^{n-1} \ln^3 t \, dt+\frac13\sum_{n=0}^\infty \int_0^1 t^{2n}\ln^3 t\, dt$$ $$=\frac32\sum_{n=1}^\infty \frac{1}{n^4}-2\sum_{n=0}^\infty \frac{1}{(2n+1)^4}=-\frac38 \zeta(4)=-\frac{\pi^4}{240}$$

Answer 3

Show that \begin{eqnarray*} I_2=\int_0^1 \frac{\ln(1-x) (\ln(1+x))^2}{x} dx = -\frac{\pi^4}{240}. \end{eqnarray*}

It is easy to show (sub $y=1-x$, geometrically expand the denominator and integrate term by term) \begin{eqnarray*} I_0=\int_0^1 \frac{(\ln(1-x))^3 }{x} dx = -6\zeta(4)= -\frac{\pi^4}{15}. \end{eqnarray*}

Now look at Ali's answer here to see that: \begin{eqnarray*} I_1 =\int_0^1 \frac{(\ln(1-x))^2 \ln(1+x)}{x} dx = -\frac{5}{8}\zeta(4)+2 \left( \operatorname{Li_4}\left(\frac12\right)+\frac78\ln2\zeta(3)-\frac14(\ln 2)^2\zeta(2)+\frac{1}{24} (\ln 2)^4 \right). \end{eqnarray*}

Next use Wolfy to get: \begin{eqnarray*} I_3 =\int_0^1 \frac{ (\ln(1+x))^3}{x} dx = 6\zeta(4)-6 \left( \operatorname{Li_4}\left(\frac12\right)+\frac{7}{8}\ln2\zeta(3)-\frac14(\ln 2)^2\zeta(2)+\frac{1}{24} (\ln 2)^4 \right). \end{eqnarray*}

Finally consider the following linear combination of the above integrals \begin{eqnarray*} I_0+3I_1+3I_2+I_3= \int_0^1 \frac{(\ln(1-x^2))^3 }{x} dx = \frac{1}{2} I_0. \end{eqnarray*} Now just do the linear algebra & we are done.