# Evaluate $\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$

Evaluate $$\int_0^1\ln(1-x)\ln x\ln(1+x) \mathrm{dx}$$

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I get (and verified by Mathematica) $$\int_0^1 \log(1-x) \log x \log(1+x) \, dx = -6 + 4 \log 2 - \log^2 2 + \frac{5}{2} \zeta(2) - 3\zeta(2) \log 2 + \frac{21}{8} \zeta(3).$$

Transforming to a double sum.

As in Marvis's answer, take \begin{align*} \log(1+x)\log(1-x) & = \left(\sum_{k=1}^{\infty} \dfrac{(-x)^k}k \right)\left(\sum_{k=1}^{\infty} \dfrac{x^k}k \right)\\ & = \sum_{k=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{(-1)^k x^{k+m}}{km}. \end{align*} Using the integral $$\int_0^1 x^r \log x \, dx = - \frac{1}{(1+r)^2},$$ we have $$\int_0^1 \log(1-x) \log x \log(1+x) \, dx = \sum_{k=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{k+1}}{km(k+m+1)^2}.$$

Evaluating the inner sum.

Partial fractions decomposition on the summand in $m$ yields \begin{align*} \sum_{m=1}^{\infty} \frac{1}{m(k+m+1)^2} &= \sum_{m=1}^{\infty} \left(\frac{1}{(k+1)^2 m} - \frac{1}{(k+1)^2 (m+k+1)} - \frac{1}{(k+1) (m+k+1)^2}\right) \\ &= \sum_{m=1}^{k+1} \frac{1}{(k+1)^2 m} - \sum_{m=k+2}^{\infty} \frac{1}{(k+1)m^2} \\ &= \frac{H_{k+1}}{(k+1)^2} - \frac{\zeta(2)}{k+1} + \frac{H^{(2)}_{k+1}}{k+1}, \end{align*} where $H^{(r)}_n = \sum_{i=1}^n i^{-r}$, the $n$th $r$-harmonic number.

Now we're left with evaluating $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{k(k+1)^2} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1} \zeta(2)}{k(k+1)} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{k(k+1)}. \tag{1}$$ We take the three sums in Eq. (1) in turn.

The first sum in Eq. (1).

For the first sum, applying partial fractions decomposition yields \begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{k(k+1)^2} &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{k} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{(k+1)} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{(k+1)^2} \\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} - 1 + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} - 1 \\ &= -2 + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)} + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} \\ &= -3 + 2 \log 2 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2}, \end{align*} where in the last step we used the evaluation for the second sum we're about to do.

The second sum in Eq. (1).

For the second sum in Equation (1), applying partial fractions decomposition yields \begin{align*} \zeta(2) \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)} &= \zeta(2) \left(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} + \sum_{k=1}^{\infty}\frac{(-1)^{k+2}}{k+1}\right) \\ &= \zeta(2) \left(2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} -1\right) \\ &= 2 \zeta(2) \log 2 - \zeta(2). \end{align*}

The third sum in Eq. (1).

For the third sum in Equation (1), applying partial fractions decomposition yields \begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{k(k+1)} &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{k} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{(k+1)} \\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)^2} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} - 1 \\ &= -1 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+1} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(k+1)^2} \\ &= -1 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} + 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} - 1 + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} - 1\\ &= -3 + 2 \log 2 + \frac{1}{2} \zeta(2) + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} ,\\ \end{align*} where in the last step we use the identity, for $p > 1$, $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^p} = (1 - 2^{1-p}) \zeta(p)$$

Combining the results.

Putting all this together, we have that the integral evaluates to $$-6 + 4 \log 2 + \frac{3}{2} \zeta(2) - 2 \zeta(2) \log 2 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k}.$$

Evaluating the alternating Euler sums.

The three remaining sums all have similar forms. Letting $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$ we have $A(1,1)$, $A(1,2)$, and $A(2,1)$ left to evaluate. Sums of the form $A(p,q)$ are known as alternating Euler sums. Euler sums and their close variants are an ongoing research area. These three, though, seem like the three simplest alternating Euler sums, and one would think that there would be relatively simple ways to evaluate them. However, despite a decent amount of work today I could not find simple proofs for any of them, either on my own or in the literature.

At any rate, we have $A(1,1) = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$, $A(1,2) = \frac{5}{8} \zeta(3)$, and $A(2,1) = \zeta(3) - \frac{1}{2}\zeta(2) \log 2$.

(Here's a link to my question where I ask for a proof for the $A(1,1)$ evaluation, and Marvis gives nice derivations for the $A(1,1)$, $A(1,2)$, and $A(2,1)$ formulas.)

Finally.

Therefore, $$\int_0^1 \log(1-x) \log x \log(1+x) \, dx = -6 + 4 \log 2 - \log^2 2 + \frac{5}{2} \zeta(2) - 3\zeta(2) \log 2 + \frac{21}{8} \zeta(3).$$

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 My answer at the end agrees with the Mathematica output given by @nbubis. – Mike Spivey Jan 11 at 5:50 I definitely love this answer! Great answer! Thank you! – Chris's wise sister Jan 11 at 7:16 You can find some ideas for evaluating those alternating Euler sums in my recent posting, though it is not as powerful as the answer by Marvis. – sos440 Jan 11 at 18:03 brilliant! this should be taught in textbooks. – nbubis Jan 11 at 18:06 @sos440: That's a nice post. It's interesting that you take the derivation to the same place as in my answer here; i.e., up to the point of needing to evaluate the alternating Euler sums. I like that you give a detailed discussion of evaluating those, too. I'll have to take a closer look at the details later. In fact, I think I'm going to print out that entire document you link to; it all looks interesting. – Mike Spivey Jan 11 at 18:11
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\begin{align} \log(1+x)\log(1-x) & = \left(\sum_{k=1}^{\infty} \dfrac{(-x)^k}k \right)\left(\sum_{k=1}^{\infty} \dfrac{x^k}k \right)\\ & = \sum_{k=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{(-1)^k x^{k+m}}{km}\\ & = \sum_{l=2}^{\infty} \left(\sum_{k=1}^{l-1} \dfrac{(-1)^k}{k(l-k)} \right) x^l \\ & = \sum_{l=1}^{\infty} \left(\sum_{k=1}^{2l-1} \dfrac{(-1)^k}{k(2l-k)} \right) x^{2l} \end{align} Now $$\int_0^1 x^{2l} \log(x) dx = -\dfrac1{(2l+1)^2}$$ Hence, the integral is $$\sum_{l=2}^{\infty} \left(\sum_{k=1}^{l-1} \dfrac{(-1)^k}{k(l-k)} \right) x^l = \sum_{l=1}^{\infty} \left(\sum_{k=1}^{2l-1} \dfrac{(-1)^{k-1}}{k(2l-k)} \right)\dfrac1{(2l+1)^2}$$

P.S: Bit too long for a comment.

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 I think you're missing a minus sign in the first line. – Ron Gordon Jan 9 at 21:42 @rlgordonma I think it is fine. $(-1)^{k-1} \times (-1) = (-1)^k$ – user17762 Jan 9 at 21:44 Ah - my mistake. Thanks. – Ron Gordon Jan 9 at 21:48 It could be a starting point. Thanks. (+1) – Chris's wise sister Jan 9 at 21:53

Mathematica gives for the final result: $$\frac{21 \zeta (3)}{8}-6-\log ^2(2)+\log (16)-\frac{1}{12} \pi ^2 (\log(64)-5)$$ The indefinite integral is really really long..

Regarding a series what's wrong with a Maclaurin series around, say, $x=1/e$?

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 Your best bet is to do a Maclurin series for $\log{(1-x)} \log{(1+x)}$ and perform the resulting integrals to get an infinite series which, I guess, comes out to what you wrote. I have evaluated $\int_0^1 dx \: \log{(1-x)} \log{x} = 2 - \frac{\pi^2}{6}$ in a similar manner. – Ron Gordon Jan 9 at 21:35 @nbubis: are you referring to do a Maclaurin series for $\log{(1-x)} \log{(1+x)}$? – Chris's wise sister Jan 9 at 21:57 @Chris'ssister - no, for $\ln(1−x)\ln(x)\ln(1+x)$. – nbubis Jan 9 at 21:59 @nbubis: OK. I'll exploit this possibility. Thanks for your suggestion (+1) – Chris's wise sister Jan 9 at 22:09