Assuming one exists, and I think it does, find a closed form for:

$$\displaystyle \int_{0}^{1}\log(1+x)\log(1-x^{3})dx$$

From it, I did manage to derive:

$$\displaystyle -\gamma+2\gamma\log(2)-\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi\left(\frac{n+4}{3}\right)}{n(n+1)}$$

But, now I am stuck on the sum.

By merely switching the signs, an integral I found was:

$$\displaystyle\int_{0}^{1}\log(1-x)\log(1+x^{3})dx=-1/2\psi_{1}(1/3)+\log^{2}(2)-2\log(2)+\frac{5\pi^{2}}{36}-\frac{\pi}{\sqrt{3}}+6$$

But, the other way around appears to be more difficult for some reason.

This is probably not new to some, but while playing around with this I did manage to find various fun identities, such as:

$$\int_{0}^{1}x^{2n}\log(1+x)dx=\frac{1}{2(3n-1)}\left(H_{6n-2}-H_{3n-1}\right)$$ when $n$ is odd and

$$\int_{0}^{1}x^{2n}\log(1+x)dx=\frac{2\log(2)}{6n+1}-\frac{1}{6n+1}\left(H_{6n+1}-H_{3n}\right)$$ when $n$ is even.

• I get $$\ln^22 - \frac18\ln^23 + 2\ln2\ln3 - \frac32\ln3 - 6\ln2 + 6-\frac{\pi}{4\sqrt{3}}(2 + \ln3) - \frac{37\pi^2}{72} + \frac12\psi_1\left(\frac13 \right ) - \frac14\text{Li}_2\left( -\frac13 \right ) + \sqrt{3}\Im\text{Li}_2\left( -\frac{i}{\sqrt{3}} \right )$$ I will typeset an answer as soon as i can – nospoon Nov 23 '15 at 14:51
• It might be not the most elegant way, but it works: Write $\log(1-x^3)=\sum_{i=1}^3\log(x-x_i)$ where $x_i$ are the roots of the equation $x^3-1=0$. The integral becomes $$\sum_{i=1}^3\int_0^1\log(x-x_i)\log(1+x)dx$$ which can be rexpressed in terms of dilogarithms quite straightforwrdly. – tired Nov 23 '15 at 15:32
• Mathematica returns this expression. – Lucian Nov 23 '15 at 17:22
• Thanks for the fast responses. Appreciate it:). I would enjoy seeing your work when and if you find time, nospoon. I tried your idea, tired. I have to admit, I don't see where it is that 'straightforward' but it does indeed work, as you said. I have not put all of the pieces completely together yet, but I have it written in terms of various Euler sums I can handle. I would like to see how you handles it if you want to outline your work. We get two for the price of one this way because $\Re\int_{0}^{1}\log(1+x)\log(1-xe^{2\pi i/3})dx=\Re\int_{0}^{1}\log(1+x)\log(1-xe^{4\pi i/3})dx$. – Cody Nov 24 '15 at 20:31
• Well, fellas, I did indeed make some progress. Albeit tedious, but doable. Just picking at it as I get time. Here is one I found: $$\sum_{n=1}^{\infty}\frac{H_{2n}}{2n+1}\cos\left(2\pi n/3\right)=-1/32\log^{2}(3)+1/48\log(3)(5\pi\sqrt(3)+54)-\frac{\pi\sqrt{3}}{2}\log(2)-\frac{31\pi^{2}}{288}+\frac{3\pi\sqrt{3}}{8}-3+\Re\left[e^{2\pi i/3}\left(3Li_{2}\left(\frac{1-e^{\pi i/3}}{2}\right)-\frac{\pi^{2}}{4}+3/2\log^{2}(2)\right)\right]$$ – Cody Nov 26 '15 at 19:41

Making the substitution $x \mapsto \frac{1-x}{1+x}$ gives $$I=\int_0^1 \ln(1+x)\ln(1-x^3)dx=2\int_0^1 \ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right)\frac{dx}{(1+x)^2}.$$

Now, we separate the integral into 7 pieces, of which only one is not elementary (containing dilogarithms): $$\ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right) \\\small =\ln^22-\ln2\ln x+\ln2\ln(3+x^2)-4\ln2\ln(1+x)-\ln x\ln(1+x)+3\ln^2(1+x)-\ln(1+x)\ln(3+x^2).$$

$1$st integral

$$\int_0^1 \frac{dx}{(1+x)^2}=-\frac1{1+x}\Bigg{|}_0^1=\frac12.$$

$2$nd integral

$$\int_0^1 \frac{\ln x}{(1+x)^2}dx=\sum_{n\geq1} \frac{(-1)^n}{n}=-\ln2.$$

$3$rd integral

$$\int_0^1 \frac{\ln(1+x)}{(1+x)^2}dx=-\frac{\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac{dx}{(1+x)^2}dx=\frac12-\frac{\ln2}{2}.$$

$4$th integral

$$\int_0^1 \frac{\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2x}{(1+x)(3+x^2)}dx \\=-\ln2+\ln3+2\operatorname{Re} \int_0^1 \frac{dx}{(1+x)(x+i\sqrt{3})}=-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}.$$

$5$th integral

$$\int_0^1 \frac{\ln x \ln(1+x)}{(1+x)^2}dx=-\frac{\ln x\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac1{1+x}\left(\frac{\ln(1+x)}{x}+\frac{\ln x}{1+x}\right)dx \\=-\ln2+\int_0^1 \frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx=\frac{\pi^2}{12}-\frac12\ln^22-\ln2.$$

$6$th integral

$$\int_0^1 \frac{\ln^2(1+x)}{(1+x)^2}dx=-\frac{\ln^2(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2\ln(1+x)}{(1+x)^2}dx \\=1-\ln2-\frac12\ln^22.$$

$7$th integral

$$\int_0^1 \frac{\ln(1+x)\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(1+x)\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac1{1+x}\left(\frac{2x\ln(1+x)}{3+x^2}+\frac{\ln(3+x^2)}{1+x}\right)dx \\=-\ln^22-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}+2J$$

where $$J=\int_0^1\frac{x\ln(1+x)}{(1+x)(3+x^2)}dx=\operatorname{Re} \int_0^1 \frac{\ln(1+x)}{(1+x)(x+i\sqrt{3})}dx \\=-\frac18\ln^22-\operatorname{Re} \,\frac1{i\sqrt{3}-1}\int_0^1 \frac{\ln(1+x)}{x+i\sqrt{3}}dx$$

Now, it is not hard to prove that $$\int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2 \ln\frac{a+1}{a-1}+\operatorname{Li}_2\left(\frac2{1-a}\right)-\operatorname{Li}_2\left(\frac1{1-a}\right).$$

Putting $a=i\sqrt{3}$ then gives, after taking the real part, and assuming the principle value of the logarithm, $$\small J=-\frac18\ln^22+\frac{\pi\sqrt{3}}{12}\ln2+\frac14\operatorname{Re}\text{Li}_2(e^{i\pi/3})-\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(e^{i\pi/3})-\frac14\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})+\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})$$

Now we can "simplify" a bit: we have $\displaystyle \,\, \operatorname{Re}\text{Li}_2(e^{i\pi/3})=\frac{\pi^2}{36}$

and $\displaystyle \,\, \operatorname{Im}\text{Li}_2(e^{i\pi/3})=\frac1{2\sqrt{3}}\psi_1\left(\frac13\right)-\frac{\pi^2}{3\sqrt{3}}.$

Furthermore, using known dilogarithm identities we have $\displaystyle \,\,\text{Li}_2(\frac12 e^{i\pi/3})=-\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\frac12\ln^2\left(\frac34-i\frac{\sqrt{3}}{4}\right),$

and since $\displaystyle \,\, \operatorname{Re}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)=\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)^2\, 3^n}=\frac14\text{Li}_2\left(-\frac13\right),$

we have $$\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi^2}{72}-\frac18\ln^2\left(\frac34\right)-\frac14\text{Li}_2\left(-\frac13\right)$$

and $$\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi}{12}\ln\left(\frac34\right)-\operatorname{Im}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right).$$

Finally, putting everything together gives the expression I posted as a comment: $$I=\ln^22 - \frac18\ln^23 + 2\ln2\ln3 - \frac32\ln3 - 6\ln2 + 6-\frac{\pi}{4\sqrt{3}}(2 + \ln3) - \frac{37\pi^2}{72} + \frac12\psi_1\left(\frac13 \right ) - \frac14\text{Li}_2\left( -\frac13 \right ) + \sqrt{3}\Im\text{Li}_2\left( -\frac{i}{\sqrt{3}} \right )$$

I really hope that the dilogarithms can be simplified some more, but I can't see how for now.

• Wow, thanks nospoon. Very nice work. There are lots of polylog identities. Maybe reference to Lewin's book we can find a way to simplify them. I managed to discover a solution, but it is not as efficient as yours. As mentioned above, I used a bunch of Euler sums. – Cody Dec 3 '15 at 11:14
• Hey nospoon, here is another integral I found if you're interested. I think it can be done the way you done the last one. But, I found: $$\int_{0}^{1}\log(1-x^{3})\log(1+x^{3})dx=1/4\left(-72+2 \gamma^{2}+\pi^{2}+4 \gamma \psi(1/6)+2\psi(1/6)-2\psi_{1}(7/6)\right)-\frac{1}{2}\left(3\sqrt{3}\pi +\frac{\pi^{2}}{6}+9\log(3)+12\log(2)-36\right)-\sum_{n=1}^{\infty}\frac{H_{2n}}{(n(6n+1))}$$. I got stuck on that last sum....so far. – Cody Dec 3 '15 at 12:14

I used your suggestion, tired, and it led to some fun results involving Euler- trig sums and polylogs. I am confident that the use of these will eventually lead to the solution to the integral listed above as well as other integral(s).

I list several I derived using the famous identities for $H_{n}$. I was mainly interested in the real parts due to the cos in the sums because they pertain to the integral in question. If anyone knows of any other closed forms, please feel free. i.e. the trilog in the 6th identity from the top or the dilog in the last one.

$$Li_{2}\left(\sqrt{3}e^{\pm \pi i/6}\right)=\frac{\pi^{2}}{9}\pm \left(2/3Cl_{2}(\frac{\pi}{3})+\frac{\pi}{3}\log(3)\right)i$$

$$\int_{0}^{1}x^{n}\log(1+x)dx=\frac{1}{2n}\left[H_{2n}-H_{n}\right], \;\ \text{n odd}$$

$$\int_{0}^{1}x^{n}\log(1+x)dx=\frac{1}{2n+1}\left(2\log(2)+H_{2n+1}-H_{n}\right), \;\ \text{n even}$$

$$\sum_{n=0}^{\infty}\frac{\cos(2\pi n/3)}{(2n+1)^{2}}=\frac{5\sqrt{3}}{12}Cl_{2}(\frac{\pi}{3})+\frac{\pi^{2}}{48}$$

$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n}x^{2n}=-\log(1+x)\log(1-x)+\sum_{n=1}^{\infty}\frac{H_{n}}{n}x^{2n}-\frac{1}{2}Li_{2}(x^{2})$$

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n^{2}}\cos(\frac{2\pi n}{3})=\frac{\pi^{2}}{18}\log(3)-\frac{\pi}{9}Cl_{2}(\frac{\pi}{3})+\frac{5}{9}\zeta(3)-\Re\left[Li_{3}\left(\sqrt{3}e^{-\pi i/6}\right)\right]$$

$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n}\cos(\frac{2\pi n}{3})=\frac{1}{8}\log^{2}(3)-\frac{7}{72}\pi^{2}$$

$$\sum_{n=1}^{\infty}\frac{H_{n}}{n}\cos(\frac{2\pi n}{3})=\frac{1}{8}\log^{2}(3)-\frac{5}{72}\pi^{2}$$

$$\sum_{n=1}^{\infty}\frac{H_{n}}{2n+1}\cos(\frac{2\pi n}{3})=\frac{-1}{4}\log^{2}(2)+\frac{5}{288}\pi^{2}-\frac{1}{32}\log^{2}(3)-\frac{\pi\sqrt{3}}{6}\log(2)+\frac{\pi\sqrt{3}}{48}\log(3)+\Re\left[e^{2\pi i/3}Li_{2}\left(\frac{\sqrt{3}}{2}e^{\pi i/6}\right)\right]$$

I have not gotten around to deriving $$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^{2}}\cos(\frac{2\pi n}{3})$$.

EDIT:

$$\sum_{n=1}^{\infty}\frac{H_{2n}}{n^{2}}\cos\left(\frac{2\pi n}{3}\right)=\frac{17}{18}\zeta(3)+\frac{7\pi}{9}Cl_{2}(\frac{\pi}{3})+\frac{5\pi^{2}}{18}\log(3)+2 \Re\left[Li_{3}\left(\sqrt{3}e^{\pi i/6}\right)\right]+\pi \Im\left[Li_{2}(-\sqrt{3}i)-Li_{2}(\sqrt{3}i)\right]-\Re\left[Li_{3}\left(\sqrt{3}e^{-\pi i/6}\right)\right]$$

Perhaps there is none, but if anyone does know of a closed form for any of those polylogs, please contribute.