# Looking for a direct way to evaluate $\int_0^1\frac{\ln(x)\ln(2+x)}{1+x}dx$

The following integral

$$I=\int_0^1\frac{\ln(x)\ln(2+x)}{1+x}dx=-\frac{13}{24}\zeta(3)$$

has been evaluated in different ways ( see here, here, here, and here) but these four solutions involve many arguments of polylogs and much simplifications are required to reach the final closed form. I am wondering if there is a simpler direct way to prove it. I tried some tricks but none worked out. Here is my best try:

Following @Zacky's idea, lets denote:

$$I(a)=\int_0^1\frac{\ln(x)\ln(1+a(1+x))}{1+x}dx$$

and note that $$I(1)=I$$ and $$I(0)=0,$$

$$I'(a)=\int_0^1\frac{\ln(x)}{1+a(1+x)}dx=\frac{\text{Li}_2\left(-\frac{a}{1+a}\right)}{a}.$$

Integrate both sides from $$a=0$$ to $$1$$,

$$\int_0^1 I'(a)=I(a)|_0^1=I(1)-I(0)=I-0=I=\int_0^1\frac{\text{Li}_2\left(-\frac{a}{1+a}\right)}{a}da$$

Integrate by parts,

$$I=\int_0^1\frac{\ln(a)\ln(1+2a)}{a(1+a)}da-\int_0^1\frac{\ln(a)\ln(1+a)}{a(1+a)}da$$

$$=\int_0^1\frac{\ln(x)\ln(1+2x)}{x(1+x)}dx+\frac{5}{8}\zeta(3).$$

To finish the proof, we need to show

$$\int_0^1\frac{\ln(x)\ln(1+2x)}{x(1+x)}dx=-\frac76\zeta(3)$$

and i am stuck here.

I also added $$I$$ to both sides:

$$2I=\int_0^1\frac{\ln(x)\ln\left(\frac{2+x}{1+2x}\right)}{1+x}dx+\int_0^1\frac{\ln(x)\ln(1+2x)}{x}dx$$

then used the subbing $$x\to (1-x)/(1+x)$$ for the first integral:

$$2I=\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{3+x}{3-x}\right)}{1+x}dx+\int_0^1\frac{\ln(x)\ln(1+2x)}{x}dx$$

and I made it even worse.

Any other thoughts ( without complicated arguments of polylogs) ?

thanks,

Bonus: Let's use the relation we obtained above:

$$I=\int_0^1\frac{\ln(x)\ln(1+2x)}{x(1+x)}dx+\frac{5}{8}\zeta(3)$$

$$I=\int_0^\infty\frac{\ln(x)\ln(1+2x)}{x(1+x)}dx-\underbrace{\int_1^\infty\frac{\ln(x)\ln(1+2x)}{x(1+x)}dx}_{x\to 1/x}+\frac{5}{8}\zeta(3)$$

$$I=\int_0^\infty\frac{\ln(x)\ln(1+2x)}{x(1+x)}dx+I-\int_0^1\frac{\ln^2(x)}{1+x}dx+\frac{5}{8}\zeta(3)$$

$$\Longrightarrow \int_0^\infty\frac{\ln(x)\ln(1+2x)}{x(1+x)}dx=\int_0^1\frac{\ln^2(x)}{1+x}dx-\frac{5}{8}\zeta(3)=\frac32\zeta(3)-\frac58\zeta(3)=\frac78\zeta(3).$$

• Use dominated convergence theorem and the fact that series expansion of $\frac{1}{1+x}$ and $\ln(1+x)$ which is in radius of convergence for the required integral Feb 12, 2022 at 4:05

On one hand, one may rewrite the integral as: \begin{align} I&=\frac{1}{2}\underbrace{\int_{0}^{1}\frac{\log^2(x)}{1+x}dx}_{\frac{3}{2}\zeta(3)}+\frac{1}{2}\underbrace{\int_{0}^{1}\frac{\log^2(2+x)}{1+x}dx}_{\frac{1}{x+2}\rightarrow y}-\frac{1}{2}\underbrace{\int_{0}^{1}\frac{\log^2\left(\frac{x}{x+2}\right)}{1+x}dx}_{\frac{x}{x+2}\rightarrow y}\\&=\frac{3}{4}\zeta(3)+\frac{1}{2}\int_{1/3}^{1/2}\frac{\log^2(y)}{y(1-y)}dy-\int_{0}^{1/3}\frac{\log^2(y)}{1-y^2}dy\\&=\frac{3}{4}\zeta(3)+\frac{1}{2}\underbrace{\int_{1/3}^{1/2}\frac{\log^2(y)}{y}dy}_{\frac{\log^3(3)-\log^3(2)}{3}}+\frac{1}{2}\left(\underbrace{\int_{0}^{1/2}\frac{\log^2(y)}{1-y}dy}_{J}-\int_{0}^{1/3}\frac{\log^2(y)}{1-y}dy\right)-\int_{0}^{1/3}\frac{\log^2(y)}{1-y^2}dy\\&=\frac{3}{4}\zeta(3)+\frac{\log^3(3)-\log^3(2)}{6}+\frac{J}{2}-\frac{1}{2}\int_{0}^{1/3}\frac{\log^2(y)}{1-y}dy-\frac{1}{2}\int_0^{1/3}\left(\frac{\log^2(y)}{1+y}+\frac{\log^2(y)}{1-y}\right)dy\\&=\frac{3}{4}\zeta(3)+\frac{\log^3(3)-\log^3(2)}{6}+\frac{J}{2}-\frac{1}{2}\underbrace{\int_0^{1/3}\left(\frac{\log^2(y)}{1+y}+\frac{2\log^2(y)}{1-y}\right)dy}_{K}\\&=\frac{3}{4}\zeta(3)+\frac{\log^3(3)-\log^3(2)}{6}+\frac{J}{2}-\frac{K}{2}\end{align}

On the other hand, the integral can also be rewritten as: \begin{align} I&=\frac{1}{4}\underbrace{\int_0^1\frac{\log^2[x(x+2)]}{1+x}dx}_{x^2+2x\rightarrow y}-\frac{1}{4}\underbrace{\int_0^1\frac{\log^2\left(\frac{x}{x+2}\right)}{1+x}dx}_{\frac{x}{x+2}\rightarrow y}\\&=\frac{1}{8}\int_{0}^{3}\frac{\log^2(y)}{1+y}dy-\frac{1}{2}\int_0^{1/3}\frac{\log^2(y)}{1-y^2}dy\\&=\frac{1}{8}\left(\int_{0}^{1/3}\frac{\log^2(y)}{1+y}dy+\underbrace{\int_{1/3}^{3}\frac{\log^2(y)}{1+y}dy}_{\frac{\log^3(3)}{3}}\right)-\frac{1}{4}\left(\int_{0}^{1/3}\frac{\log^2(y)}{1+y}dy+\int_{0}^{1/3}\frac{\log^2(y)}{1-y}dy\right)\\&=\frac{\log^3(3)}{24}-\frac{K}{8}\end{align}

So it boils down to the computation of $$J$$. Since the OP wants an answer without resorting too heavily into polylogarithm functions, let's evaluate $$J$$ as the following: \begin{align}J&=\underbrace{\int_{0}^{1/2}\frac{\log^2(y)}{1-y}dy}_{y\rightarrow \frac{x}{1+x}}=\int_0^1\frac{\log^2\left(\frac{x}{1+x}\right)}{1+x}dx\\&=\underbrace{\int_0^1\frac{\log^2(x)+\log^2(1+x)}{1+x}dx}_{\frac{3}{2}\zeta(3)+\frac{\log^3(2)}{3}}-2\underbrace{\int_0^1\frac{\log(x)\log(1+x)}{1+x}dx}_{IBP}\\&=\frac{3}{2}\zeta(3)+\frac{\log^3(2)}{3}+\underbrace{\int_0^1\frac{\log^2(1+x)}{x}dx}_{\frac{1}{1+x}\rightarrow y}\\&=\frac{3}{2}\zeta(3)+\frac{\log^3(2)}{3}+\int_{1/2}^{1}\frac{\log^2(y)}{y(1-y)}dy=\frac{3}{2}\zeta(3)+\frac{2}{3}\log^3(2)+\int_{1/2}^{1}\frac{\log^2(y)}{1-y}dy\\2J&=\frac{3}{2}\zeta(3)+\frac{2}{3}\log^3(2)+\left(\int_{0}^{1/2}\frac{\log^2(y)}{1-y}dy+\int_{1/2}^{1}\frac{\log^2(y)}{1-y}dy\right)\\2J&=\frac{3}{2}\zeta(3)+\frac{2}{3}\log^3(2)+\underbrace{\int_{0}^{1}\frac{\log^2(y)}{1-y}dy}_{2\zeta(3)}\end{align} $$\boxed{J=\frac{7}{4}\zeta(3)+\frac{\log^3(2)}{3}}$$

Equating both representations of $$I$$: $$\frac{3}{4}\zeta(3)+\frac{\log^3(3)-\log^3(2)}{6}+\frac{7}{8}\zeta(3)+\frac{\log^3(2)}{6}-\frac{K}{2}=\frac{\log^3(2)}{24}-\frac{K}{8}$$ $$\boxed{K=\frac{13}{3}\zeta(3)+\frac{\log^3(3)}{3}}$$

Hence: $$I=\frac{\log^3(3)}{24}-\frac{1}{8}\left(\frac{13}{3}\zeta(3)+\frac{\log^3(3)}{3}\right)=-\frac{13}{24}\zeta(3)$$

• (+1) Very nice .. thank you Jun 26, 2022 at 4:33
• Thank you Ali! Coming from you it means a lot to me Jun 26, 2022 at 5:32
• This is the solution i was searching for a while ! Thank you !
– FDP
Jun 28, 2022 at 8:48
• Thank you for the positive feedback @FDP Jun 29, 2022 at 2:03
• You're welcome!. I'm wondering if one can simplify your solution.
– FDP
Jun 29, 2022 at 7:10