Well, I've been trying to solve the following integral: \begin{equation*} \int_0^1\frac{\text{Li}_3(x)}{1+x}\mathrm dx, \end{equation*} where by integration by parts, making $u=\text{Li}_3(x)$ and $\mathrm dv=\frac1{1+x}$, i got: \begin{align*} \int_0^1\frac{\text{Li}_3(x)}{1+x}\mathrm dx=\ln(2)\zeta(3)-\int_0^1\frac{\text{Li}_2(x)\ln(1+x)}{x}\mathrm dx.\tag{1} \end{align*} Particularly, I was curious about the integral on the right side of (1), so I proceeded naively to apply integration by parts again. This time, doing $u=\text{Li}_2(x)\ln(1+x)$ and $\mathrm dv=\frac1x$: (note: the choice $u=\text{Li}_2(x)$ and $\mathrm dv=\frac{\ln(1+x)}{x}$ it has a certain symmetry with the integral I want to calculate) $$\int_0^1\frac{\text{Li}_2(x)\ln(1+x)}{x}\mathrm dx=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+x)}{x}\mathrm dx-\int_0^1\frac{\text{Li}_2(x)\ln(x)}{1+x}\mathrm dx.$$ In order not to extend this post, I briefly know that it is possible to calculate the first integral above, but the second is another problem I don't know how to deal with. There is the possibility of using generator functions for the integral of my question, but I'm trying to avoid that, another thing is that I can still accept this type of solution.
1 Answers
In the comments, @Svyatoslav found
$$\int_0^1\frac{\text{Li}_3(x)}{1+x}dx=-\frac{1}{2}\int_0^1\frac{\ln^2x\ln(1-x)}{x}dx-\frac{1}{2}\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx-\frac{\ln2}{2}\int_0^1\frac{\ln^2x}{1+x}dx.$$
The first and third integrals are trivial. For the second one, apply IBP:
$$\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx=\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}dx-2\int_0^1\frac{\ln x\ln(1-x)\ln(1+x)}{x}dx.$$
The second integral is well known. For the first one, we follow Cornel's method:
$$I=\int_0^1\frac{\ln^2x}{1-x}\ln\left(\frac{2}{1+x}\right)dx=\int_0^1\frac{\ln^2x}{1-x}\left(\int_x^1\frac{dy}{1+y}\right)dx$$
$$=\int_0^1\int_0^y\frac{\ln^2x}{(1+y)(1-x)}dxdy\overset{x=yz}{=}\int_0^1\int_0^1\frac{y\ln^2(yz)}{(1+y)(1-yz)}dzdy$$
replace $z$ by $x$ then add the integral $I$ to both sides,
$$2I=\int_0^1\int_0^1\frac{y\ln^2(xy)}{(1+y)(1-xy)}dxdy+\int_0^1\int_0^1\frac{y\ln^2(xy)}{(1+y)(1-xy)}dxdy$$
swap $x$ and $y$ in the second integral,
$$2I=\int_0^1\int_0^1\frac{y\ln^2(xy)}{(1+y)(1-xy)}dxdy+\int_0^1\int_0^1\frac{x\ln^2(xy)}{(1+x)(1-xy)}dxdy$$
$$=\int_0^1\int_0^1\frac{[(1+x)(1+y)-(1-xy)]\ln^2(xy)}{(1+x)(1+y)(1-xy)}dxdy$$
$$=\int_0^1\int_0^1\frac{\ln^2(xy)}{1-xy}dxdy-\int_0^1\int_0^1\frac{\ln^2(xy)}{(1+x)(1+y)}dxdy$$
The first integral:
$$\int_0^1\int_0^1\frac{\ln^2(xy)}{1-xy}dxdy\overset{x=t/y}{=}\int_0^1\int_0^y\frac{\ln^2t}{y(1-t)}dtdy$$
$$=\int_0^1\frac{\ln^2t}{1-t}\left(\int_t^1\frac{dy}{y}\right)dt=-\int_0^1\frac{\ln^3t}{1-t}dt=6\zeta(4).$$
The second integral:
$$\int_0^1\int_0^1\frac{\ln^2(xy)}{(1+x)(1+y)}dxdy=\int_0^1\int_0^1\frac{\ln^2x+2\ln x\ln y+\ln^2y}{(1+x)(1+y)}dxdy$$
$$=\int_0^1\int_0^1\frac{\ln^2x}{(1+x)(1+y)}dxdy+2\int_0^1\int_0^1\frac{\ln x\ln y}{(1+x)(1+y)}dxdy$$ $$+\int_0^1\int_0^1\frac{\ln^2y}{(1+x)(1+y)}dxdy$$
Note that the first and third integrals are equivalents, therefore
$$\int_0^1\int_0^1\frac{\ln^2(xy)}{(1+x)(1+y)}dxdy=2\left(\int_0^1\frac{\ln^2x}{1+x}dx\right)\left(\int_0^1\frac{dy}{1+y}\right)$$ $$+2\left(\int_0^1\frac{\ln x}{1+x}dx\right)\left(\int_0^1\frac{\ln y}{1+y}dy\right)$$
$$=2\left(\frac32\zeta(3)\right)(\ln2)+2\left(-\frac12\zeta(2)\right)^2$$
$$=3\ln2\zeta(3)+\frac54\zeta(4).$$
Finally,
$$I=\int_0^1\frac{\ln^2x}{1-x}\ln\left(\frac{2}{1+x}\right)dx=\frac{19}{8}\zeta(4)-\frac32\ln2\zeta(3).$$
Note that
$$I=\ln2\int_0^1\frac{\ln^2x}{1-x}dx-\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}dx=2\ln2\zeta(3)-\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}dx$$
$$\Longrightarrow \int_0^1\frac{\ln^2x\ln(1+x)}{1-x}dx=\frac72\ln2\zeta(3)-\frac{19}{8}\zeta(4).$$
- 25,498
MZIntegratepackages I have:$$\frac{\pi ^4}{60}+\frac{1}{12} \pi ^2 \ln ^2(2)-\frac{\ln ^4(2)}{ 12}-2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{3}{4} \ln (2) \zeta (3)$$ – Mariusz Iwaniuk Nov 07 '21 at 10:23