6

I want to show that $$\displaystyle I=\int_0^1\frac{\log(x)\log(1-x)}{x(1-x)}\operatorname{Li}_2(x)\mathrm{dx}=5\zeta(2)\zeta(3)-8\zeta(5)$$

I started by doing partial fraction, then we get $$\displaystyle I=\int_0^1\frac{\log(x)\log(1-x)}{x}\operatorname{Li}_2(x)\mathrm{d}x+\int_0^1\frac{\log(x)\log(1-x)}{1-x}\operatorname{Li}_2(x)\mathrm{d}x$$

Then in the second integral I did $1-x\to x$ $$\displaystyle\implies I=\int_0^1\frac{\log(x)\log(1-x)}{x}\operatorname{Li}_2(x)\mathrm{d}x+\int_0^1\frac{\log(x)\log(1-x)}{x}\operatorname{Li}_2(1-x)\mathrm{d}x$$

$$\displaystyle=\int_0^1\left(\frac{\log(x)\log(1-x)}{x}\right)(\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x))\mathrm{d}x$$

$$\displaystyle(\because \operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)=\frac{\pi^2}{6}-\log(x)\log(1-x))$$

$$\displaystyle\implies I=\frac{\pi^2}{6}\int_0^1\frac{\log(x)\log(1-x)}{x}\mathrm{d}x-\int_0^1\frac{\log^2(x)\log^2(1-x)}{x}\mathrm{d}x$$
The first integral is easy to evaluate, $\displaystyle I_1=\frac{\pi^2}{6}\int_0^1\frac{\log(x)\log(1-x)}{x}\mathrm{d}x$

$$\overset{IBP}{=}\frac{\pi^2}{6}[-\operatorname{Li}_2(x)\log(x)]_0^1+\frac{\pi^2}{6}\int_0^1\frac{\operatorname{Li}_2(x)}{x}\mathrm{d}x=\frac{\pi^2}{6}[\operatorname{Li}_3(x)]_0^1$$

$$\displaystyle=\frac{\pi^2}{6}\operatorname{Li}_3(1)=\zeta(2)\zeta(3)$$

The second integral I have no idea how to solve it, in WolframAlpha it gives an aproximation but I think the value is the same as $4\zeta(2)\zeta(3)-8\zeta(5)$, so I think everything I did is right.

Dang Dang
  • 320
bruh Jardim
  • 119
  • 2
    See below link(B_2 integral) for the second integral https://math.stackexchange.com/questions/3718397/how-to-evaluate-sum-n-1-infty-frach-n2n3?noredirect=1 – Frank Jun 18 '25 at 20:18

1 Answers1

7

$$\displaystyle I=\int_0^1\frac{\log(x)}{x}(\log(1-x))\operatorname{Li}_2(x)\mathrm{d}x+\int_0^1\frac{\log(x)}{1-x}(\log(1-x))\operatorname{Li}_2(x)\mathrm{d}x=I_1+I_2$$

Take $I_2$,

$$I_2=\int_0^1\frac{\log(x)\log(1-x)\operatorname{Li}_2(x)}{1-x}\mathrm{d}x$$

Use the below reflection formula in $I_2$,

$$\color{blue}{\operatorname{Li}_2(x)=\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(1-x)}$$

$$I_2=\zeta(2)\int_0^1\frac{\log(x)\log(1-x)}{1-x}\mathrm{d}x-\int_0^1\frac{\log^2(x)\log^2(1-x)}{1-x}\mathrm{d}x-\int_0^1\frac{\log(x)\log(1-x)\operatorname{Li}_2(1-x)}{1-x}\mathrm{d}x$$

Perform $1-x \to x$ only for the third integral,

$$=\zeta(2)\int_0^1\frac{\log(x)\log(1-x)}{1-x}\mathrm{d}x-\int_0^1\frac{\log^2(x)\log^2(1-x)}{1-x}\mathrm{d}x-\int_0^1 \frac{\log(x)}{x} \log(1 - x)\operatorname{Li}_2(x)\, dx$$

The first two integrals are trivial via the Beta function, whereas for the third integral, if you notice just hard enough, it's $I_1$

$$I_2=\zeta(2)\int_0^1\frac{\log(x)\log(1-x)}{1-x}\mathrm{d}x-\int_0^1\frac{\log^2(x)\log^2(1-x)}{1-x}\mathrm{d}x-I_1$$

The above step does lead to something nice,

$$I_1+I_2=\zeta(2)\int_0^1\frac{\log(x)\log(1-x)}{1-x}\mathrm{d}x-\int_0^1\frac{\log^2(x)\log^2(1-x)}{1-x}\mathrm{d}x=\zeta(2)A-B$$

Remember $I=I_1+I_2$

As previously mentioned these are trivial integrals using the Beta function,

Put $1-x\to x$ and use the result from here, $$A=\int_0^1\frac{\log(x)\log(1-x)}{1-x}\mathrm{d}x=\zeta(3)$$

Put $1-x\to x$ and use the modified result from here

$$B=\int_0^1\frac{\log^2(x)\log^2(1-x)}{1-x}\mathrm{d}x=8\zeta(5)-4\zeta(2)\zeta(3)$$

Therefore,

$$I=I_1+I_2=5\zeta(2)\zeta(3)-8\zeta(5)$$

Numerical checks are here and here

$$\therefore \int_0^1\frac{\log(x)\log(1-x)}{x(1-x)}\operatorname{Li}_2(x)\mathrm{d}x = 5\zeta(2)\zeta(3)-8\zeta(5)$$

I believe there's more explaining on the proof to $B$,

$$B=\int_0^1\frac{\log^2(x)\log^2(1-x)}{1-x}\mathrm{d}x\underset{1-x \to x}\implies \int_0^1\frac{\log^2(x)\log^2(1-x)}{x}\mathrm{d}x$$

$$\beta (m,n) = \int_0^1 x^{m - 1} (1 - x)^{n - 1} \, dx$$ we see that $$\lim_{m \to 0^+} \lim_{n \to 1} \partial^2_m \partial^2_n \beta (m,n) = \int_0^1 \frac{\log^2 x \log^2 (1 - x)}{x} \, dx$$

Since showing all the steps of this double derivative and referring to values will eat up lot of space,

Here's the mathematica script,

D[D[Beta[m, n], {n, 2}], {m, 2}] // 
Limit[#, m -> 0, Direction -> "FromAbove"] & // 
Limit[#, n -> 1] &

Which gives us,

$$B=\frac{1}{3} \left( \pi^2 \, \psi^{(2)}(1) - \psi^{(4)}(1) \right)=8\zeta(5)-4\zeta(2)\zeta(3)$$

Refer here

Amrut Ayan
  • 8,887