Evaluación de $\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$

Question

En este rosca

un amigo publicó la siguiente integral

$$I=\int^1_0 \frac{\log(1+x)\log(1-x) \log(x)}{x}\, \mathrm dx$$

Lo mejor que podemos hacer es expresarlo en términos de sumas de Euler

$$I=-\frac{\zeta^2(2)}{2}+ \sum_{n\geq 1}\frac{(-1)^{n-1}}{n^2} H_{n}^{(2)}+\sum_{n\geq 1}\frac{(-1)^{n-1}}{n^3}H_{n}$$

Me pregunto si el enfoque que he seguido ha complicado la integral. ¿Qué enfoque seguirías tú para resolver la integral?, ¿podemos encontrar una solución mejor?

Answer 1

\begin{align*} J&=\int_0^1 \frac{\ln x\ln(1-x)\ln(1+x)}{x}\,dx\\ &\overset{IBP}=\frac{1}{2}\Big[\ln^2 x\ln(1-x)\ln(1+x)\Big]_0^1 -\frac{1}{2}\int_0^1 \ln^2 x\left(\frac{\ln(1-x)}{1+x}-\frac{\ln(1+x)}{1-x}\right)\,dx\\ &=\frac{1}{2}\int_0^1 \ln^2 x\left(\frac{\ln(1+x)}{1-x}-\frac{\ln(1-x)}{1+x}\right)\,dx\\ K&=\int_0^1 \frac{\ln^2 x\ln(1+x)}{1-x}\,dx,L=\int_0^1 \frac{\ln^2 x\ln(1-x)}{1+x}\,dx,M=\int_0^1\frac{\ln(1+x)\ln^2 x}{1+x}\,dx\\ &\overset{IBP}=\left[\left(\int_0^x \frac{\ln^2 t}{1-t}\,dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x \frac{\ln^2 t}{1-t}\,dt\right)\,dx\\ &\overset{u(t)=xt}=2\zeta(3)\ln 2-\int_0^1 \int_0^1 \frac{x\ln^2(tx)}{(1-tx)(1+x)}\,dt\,dx\\ &=2\zeta(3)\ln 2-\frac{1}{2}\left(\int_0^1 \int_0^1 \frac{x\ln^2(tx)}{(1-tx)(1+x)}\,dt\,dx+\int_0^1 \int_0^1 \frac{t\ln^2(tx)}{(1-tx)(1+t)}\,dt\,dx\right)\\ &=2\zeta(3)\ln 2-\frac{1}{2}\left(\int_0^1 \int_0^1 \frac{\ln^2(tx)}{1-tx}\,dt\,dx-\int_0^1 \int_0^1 \frac{\ln^2(tx)}{(1+t)(1+x)}\,dt\,dx\right)\\ &=2\zeta(3)\ln 2+\int_0^1 \frac{\ln^2 x+\ln t\ln x}{(1+t)(1+x)}\,dt\,dx-\frac{1}{2}\int_0^1\int_0^1 \frac{\ln^2 (tx)}{1-tx}\,dt\,dx\\ &=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}-\frac{1}{2}\int_0^1\int_0^1 \frac{\ln^2 (tx)}{1-tx}\,dt\,dx\\ &\overset{u=tx}=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}-\frac{1}{2}\int_0^1 \frac{1}{x}\left(\int_0^x \frac{\ln^2 u}{1-u}\,du\right)\,dx\\ &\overset{IBP}=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}-\frac{1}{2}\left[\ln x\left(\int_0^x \frac{\ln^2 u}{1-u}\,du\right)\right]_0^1+\frac{1}{2}\int_0^1 \frac{\ln^3 x}{1-x}\,dx\\ &=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}+\frac{1}{2}\int_0^1 \frac{\ln^3 x}{1-x}\,dx\\ &=\frac{7}{2}\zeta(3)\ln 2+\frac{\pi^4}{144}+\frac{1}{2}\times -\frac{\pi^4}{15}\\ &=\boxed{\frac{7}{2}\zeta(3)\ln 2-\frac{19\pi^4}{720}} \end{align*} \begin{align*} 0&<A<1\\ L(A)&=\int_0^A \frac{\ln^2 x\ln(1-x)}{1+x}\,dx\\ &\overset{IBP}=\left[\left(\int_0^x \frac{\ln^2 t}{1+t}\,dt\right)\ln(1-x)\right]_0^A+\int_0^A \frac{1}{1-x}\left(\int_0^x \frac{\ln^2 t}{1+t}\,dt\right)\,dx\\ &\overset{t(u)=ux}=\left(\int_0^A \frac{\ln^2 t}{1+t}dt\right)\ln(1-A)+\int_0^A \left(\int_0^1 \frac{x\ln^2(ux)}{(1-x)(1+ux)}\,du\right)\,dx\\ &=\left(\int_0^A \frac{\ln^2 t}{1+t}\,dt\right)\ln(1-A)+\int_0^A\left(\int_0^1 \frac{\ln^2(ux)}{(1+u)(1-x)}du\right)dx-\\ &\int_0^A\left(\int_0^1 \frac{\ln^2(ux)}{(1+u)(1+ux)}du\right)dx\\ &=\left(\int_0^A \frac{\ln^2 t}{1+t}\,dt-\frac{3}{2}\zeta(3)\right)\ln(1-A)+\ln 2\int_0^A\frac{\ln^2 x}{1-x}\,dx-\frac{\pi^2}{6}\int_0^A \frac{\ln x}{1-x}\,dx-\\ &\int_0^A\left(\int_0^1 \frac{\ln^2(ux)}{(1+u)(1+ux)}du\right)dx\\ L&=\lim_{A\rightarrow 1}L(A)\\ &=2\zeta(3)\ln 2+\frac{\pi^4}{36}-\int_0^1\left(\int_0^1 \frac{\ln^2(ux)}{(1+u)(1+ux)}du\right)dx\\ &\overset{t(x)=xu}=2\zeta(3)\ln 2+\frac{\pi^4}{36}-\int_0^1\frac{1}{u(1+u)}\left(\int_0^u \frac{\ln^2 t}{1+t}\,dt\right)\,du\\ &\overset{IBP}=2\zeta(3)\ln 2+\frac{\pi^4}{36}-\left[\ln\left(\frac{u}{1+u}\right)\left(\int_0^u \frac{\ln^2 t}{1+t}dt\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{u}{1+u}\right)\ln^2 u}{1+u}du\\ &=\frac{7}{2}\zeta(3)\ln 2-\frac{11}{360}\pi^4-M\\ \end{align*} \begin{align*} U&=\int_0^1 \frac{\ln^3\left(\frac{x}{1+x}\right)}{1+x}\,dx\\ &\overset{y=\frac{x}{1+x}}=\int_0^{\frac{1}{2}}\frac{\ln^3 x}{1-x}\,dx\\ U&=\int_0^1 \frac{\ln^3 x}{1+x}\,dx-\int_0^1 \frac{\ln^3(1+x)}{1+x}\,dx-3\int_0^1 \frac{\ln^2 x\ln(1+x)}{1+x}\,dx+3\int_0^1 \frac{\ln^2(1+x)\ln x}{1+x}\,dx\\ &=\int_0^1 \frac{\ln^3 x}{1+x}\,dx-\frac{1}{4}\ln^4 2-3M+\Big[\ln^3(1+x)\ln x\Big]_0^1-\int_0^1 \frac{\ln^3(1+t)}{t}\,dt\\ &\overset{x=\frac{1}{1+t}}=\int_0^1 \frac{\ln^3 x}{1+x}\,dx-\frac{1}{4}\ln^4 2-3M+\int_{\frac{1}{2}}^1 \frac{\ln^3 x}{x(1-x)}\,dx\\ &=\int_0^1 \frac{\ln^3 x}{1+x}\,dx-\frac{1}{4}\ln^4 2-3M+\int_{\frac{1}{2}}^1 \frac{\ln^3 x}{x}\,dx-\int_{\frac{1}{2}}^1 \frac{\ln^3 x}{1-x}\,dx\\ &=2\int_0^1 \frac{\ln^3 x}{1-x^2}\,dx-\frac{1}{2}\ln^4 2-3M-\int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx\\ &=\left(2\int_0^1 \frac{\ln^3 x}{1-x}\,dx-\int_0^1 \frac{2t\ln^3 t}{1-t}\,dt\right)-\frac{1}{2}\ln^4 2-3M-\int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx\\ &\overset{x=t^2}=\frac{15}{8}\int_0^1 \frac{\ln^3 x}{1-x}\,dx-\frac{1}{2}\ln^4 2-3M-\int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx\\ \end{align*}

Por lo tanto, \begin{align*} M&=\frac{5}{8}\int_0^1 \frac{\ln^3 x}{1-x}\,dx-\frac{1}{6}\ln^4 2-\frac{2}{3}\int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx\\ \int_0^{\frac{1}{2}} \frac{\ln^3 x}{1-x}\,dx&\overset{y=2x}=\frac{1}{2}\int_0^1 \frac{\ln^3\left(\frac{1}{2}x\right)}{1-\frac{1}{2}x}\,dx\\ &=\frac{1}{2}\int_0^1 \frac{\ln^3 x}{1-\frac{1}{2}x}\,dx-\frac{\ln^3 2}{2}\int_0^1 \frac{1}{1-\frac{1}{2}x}\,dx-\\ &\frac{3\ln 2}{2}\int_0^1 \frac{\ln^2 x}{1-\frac{1}{2}x}dx+\frac{3\ln^2 2}{2}\int_0^1 \frac{\ln x}{1-\frac{1}{2}x}dx\\ &=-6\text{Li}_4\left(\frac{1}{2}\right)-\ln^4 2-6\ln 2\text{Li}_3\left(\frac{1}{2}\right)-3\ln^2 2 \text{Li}_2\left(\frac{1}{2}\right)\\ &=-6\text{Li}_4\left(\frac{1}{2}\right)-\frac{21\zeta(3)}{4}\ln 2+\frac{\pi^2 \ln^2 2 }{4}-\frac{\ln^4 2}{2}\\ M&=4\text{Li}_4\left(\frac{1}{2}\right)-\frac{\pi^4}{24}+\frac{7\zeta(3)\ln 2}{2}-\frac{\pi^2 \ln^2 2}{6}+\frac{\ln^4 2}{6}\\ L&=\boxed{\frac{\pi^4}{90}-4\text{Li}_4\left(\frac{1}{2}\right)+\frac{\pi^2 \ln^2 2}{6}-\frac{\ln^4 2}{6}}\\ J&=\frac{1}{2}\left(K-L\right)\\ &=\boxed{2\text{Li}_4\left(\frac{1}{2}\right)-\frac{3\pi^4}{160}+\frac{7\zeta(3)\ln 2}{4}-\frac{\pi^2 \ln^2 2}{12}+\frac{\ln^4 2}{12}} \end{align*}

NB: Supongo que sí, $r\geq 1,0< a\leq 1$ , números enteros \begin{align*} \int_0^1 \frac{\ln^r x }{1-ax}\,dx&=\frac{(-1)^r r!}{a}\text{Li}_{r+1}(a)\\ \text{Li}_2\left(\frac{1}{2}\right)&=\frac{\pi^2}{12}-\frac{\ln^2 2}{2},\text{Li}_2(1)=\zeta(2)=\frac{\pi^2}{6}\\ \text{Li}_3(1)&=\zeta(3),\text{Li}_3\left(\frac{1}{2}\right)=\frac{7\zeta(3)}{8}+\frac{\ln^3 2}{6}-\frac{\pi^2\ln 2}{12},\text{Li}_4(1)=\zeta(4)=\frac{\pi^4}{90} \end{align*}

Answer 2

Yo haría el siguiente cambio de variable.

$$x=e^{-t}$$ Entonces podemos representar la integral de la siguiente manera:

$$I=-\int_{0}^{\infty}t\ln(1+e^{-t})\ln(1-e^{-t})\;dt$$ Ahora, aplica la expansión de Taylor del logaritmo:

$$\ln(1+x)=\sum_{i=1}^{\infty}(-1)^{i-1}\frac{x^i}{i}$$

$$I=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^{i-1}}{ij}\int_{0}^{\infty}te^{-(i+j)t}dt=$$

$$=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{(-1)^{i-1}}{ij(i+j)^2}$$

Answer 3

$$-\frac{\partial^2}{\partial s\partial t}\left[B(s+1,t+1)\;_3 F_2(1,1,s+1;2,s+t+2;-1)\right]_{s=t=0}$$

Puede ser que la función hipergeométrica sea sumable. En este caso, la diferenciación es trivial. (B denota la función beta de Euler).