Cómo evaluar $$\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx$$
Mi intento
$$\begin{align}\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx&=\frac{1}{4}\int_{0}^{1}{\ln(1+x)}d(\ln^2(1+x^2))\\ &=\frac{1}{4}(\ln(1+x)\ln^2(1+x^2)|_0^1-\int_{0}^{1}{\ln^2(1+x^2)}d\ln(1+x))\\ &=\frac{1}{4}(\ln^32-\int_{0}^{1}\frac{\ln^2(1+x^2)}{1+x}dx)\\ \end{align}$$
$$\begin {align} &\int_{0}^{1} \frac{\ln ^{2}\left(1+x^{2}\right)}{1+x} d x\\=&\int_{0}^{1} \frac{\ln ^{2}\left(1-y^{2}\right)}{1+i y} i d y-\int_{0}^{\frac{\pi}{2}} \frac{\ln ^{2}\left(1+e^{i 2 \theta}\right)}{1+e^{i \theta}} i e^{i \theta} d\theta\\ =&\int_{0}^{1} \frac{y \ln ^{2}\left(1-y^{2}\right)}{1+y^{2}} dy+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \tan \left(\frac{\theta}{2}\right) \ln ^{2}(2 \cos \theta)d\theta+\int_{0}^{\frac{\pi}{2}} \theta \ln (2 \cos \theta)d\theta\\&-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^{2} \tan \left(\frac{\theta}{2}\right)d\theta \end {align}$$
Pero, ¿cómo evaluar $$\int_{0}^{1}\frac{\ln^2(1+x^2)}{1+x}dx$$
La respuesta de Mathematica es $$\frac{1}{96}(24\pi\mathbf{G}+\pi^2\ln2+8\ln^32-60\zeta(3))$$
$\mathbf{G}$ es el catalán es constante.
Mi respuesta
Respuestas
¿Demasiados anuncios?
Usted ya ha utilizado el contorno de integración para dividir la integral en cuatro partes: $$\begin {align} &\int_{0}^{1} \frac{\ln ^{2}\left(1+x^{2}\right)}{1+x}dx\\ =&\int_{0}^{1} \frac{x \ln ^{2}\left(1-x^{2}\right)}{1+x^{2}} dx+\frac{1}{2} \int_{0}^{\pi/2} \tan \left(\frac{x}{2}\right) \ln ^{2}(2 \cos x)dx+\int_{0}^{\pi/2} x \ln (2 \cos x)dx\\&-\frac{1}{2} \int_{0}^{\pi/2} x^{2} \tan \left(\frac{x}{2}\right)dx \end {align}$$
Indicar los cuatro integrales por $I_1,\cdots, I_4$. $$I_1 = \frac{1}{2}\int_0^1 \frac{\ln^2(1-x)}{1+x}dx = \frac{1}{2}\int_0^1 \frac{\ln^2(x)}{2-x}dx = \text{Li}_3(\frac{1}{2}) = \frac{\ln^2 3}{6}-\frac{\pi^2}{12}\ln 2 +\frac{7}{8}\zeta(3)$$ Para $I_2$, tangente a la mitad de la sustitución da $$I_2 = \int_0^1\frac{2t\ln^2(2\frac{1-t^2}{1+t^2})}{1+t^2}dt = \int_0^1 \frac{\ln^2(2\frac{1-t}{1+t})}{1+t}dt = \int_0^1 \frac{\ln^2 (2t)}{1+t}dt = \frac{3\zeta(3)}{2}+\ln^3 2 -\frac{\pi^2}{6}\ln 2$$ donde el penúltimo paso consiste en $t\mapsto (1-t)/(1+t)$.
Para $I_3$, la integración por parte da $$I_3 = \int_0^{\pi/2} (\frac{\pi}{2}-x)\ln(2\sin x) dx = -\frac{\pi^2}{8}\ln 2+\frac{1}{2}\int_0^{\pi/2} x^2 \cot x dx $$ Uso de la identidad $$\sum_{n=1}^N \sin(nx) = \frac{1}{2}\cot\frac{x}{2}-\frac{\cos(N+1/2)x}{2\sin(x/2)}$$ dejando $N\to \infty$, y el resto $\to 0$ por Riemann-Lebesuge lema, por lo que $$\int_0^{\pi/2} x^2\cot x dx = 2\sum_{n=1}^\infty \int_0^{\pi/2} x^2\sin(2nx) dx = \frac{\pi^2}{4}\ln 2 -\frac{7\zeta(3)}{8}$$ Note that each summand can be evaluated explicitly. Therefore $I_3 = -7\zeta(3)/16$.
Para $I_4$, el uso de la misma identidad, da $$I_4 = 2\sum_{n=1}^\infty \int_0^{\pi/2} x^2\sin[n(\pi-x)]dx = 2\sum_{n=1}^\infty (-1)^{n+1} \int_0^{\pi/2} x^2\sin (nx)dx$$ de la serie de la definición del catalán constante de inmediato da $$I_4 = 2\pi G - \frac{21\zeta(3)}{8} - \frac{\pi^2}{4}\ln 2$$
La combinación de todos estos deben dar el resultado.
Una alternativa "primaria" de la solución (sin contorno integral, sin número complejo en realidad)
Mantenga la respiración !
\begin{align}\text{A}&=\int_0^1 \frac{x\ln(1-x)\ln(1+x^2)}{1+x^2}\,dx\\ \text{B}&=\int_0^1 \frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}\,dx\\ \text{A+B}&=\int_0^1 \frac{x\ln(1-x^2)\ln(1+x^2)}{1+x^2}\,dx\\ \end{align}
Realizar el cambio de variable $\displaystyle y=x^2$,
\begin{align} \text{A+B}&=\frac{1}{2}\int_0^1 \frac{\ln(1-x)\ln(1+x)}{1+x}\,dx\\ &=\frac{1}{4}\int_0^1 \frac{\ln^2(1-x)}{1+x}\,dx+\frac{1}{4}\int_0^1 \frac{\ln^2(1+x)}{1+x}\,dx-\frac{1}{4}\int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}\,dx\\ &=\frac{1}{4}\int_0^1 \frac{\ln^2\left(\frac{2x}{1+x}\right)}{1+x}\,dx+\frac{1}{4}\int_0^1 \frac{\ln^2(1+x)}{1+x}\,dx-\frac{1}{4}\int_0^1 \frac{\ln^2 x }{1+x}\,dx \end{align}
En la primera y la tercera integral realizar el cambio de variable $y=\dfrac{1-x}{1+x}$,
\begin{align} \text{A+B}&=\frac{1}{4}\int_0^1 \frac{\ln^2\left(\frac{2x}{1+x}\right)}{1+x}\,dx+\frac{1}{4}\int_0^1 \frac{\ln^2(1+x)}{1+x}\,dx-\frac{1}{2}\int_0^1 \frac{\ln^2 x }{1+x}\,dx\\ &=\frac{1}{4}\int_0^1 \frac{\ln^2(1+x)}{1+x}\,dx-\frac{1}{2}\int_0^1 \frac{\ln(1+x)\ln x}{1+x}\,dx-\frac{\ln 2}{2}\int_0^1 \frac{\ln(1+x)}{1+x}\,dx\\ &+\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1+x}\,dx+\frac{\ln 2}{2}\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{\ln^2 2}{4}\int_0^1 \frac{1}{1+x}\,dx+\frac{1}{4}\int_0^1 \frac{\ln^2(1+x)}{1+x}\,dx-\\ &\frac{1}{4}\int_0^1 \frac{\ln^2 x }{1+x}\,dx\\ &=\frac{1}{6}\ln^3 2-\frac{1}{2}\int_0^1 \frac{\ln(1+x)\ln x}{1+x}\,dx+\frac{1}{2}\ln 2\int_0^1 \frac{\ln x}{1+x}\,dx \end{align}
\begin{align} \text{A-B}&=\int_0^1 \frac{x\ln\left(\frac{1-x}{1+x}\right)\ln(1+x^2)}{1+x^2}\,dx\\ \end{align}
Realizar el cambio de variable $y=\dfrac{1-x}{1+x}$
\begin{align} \text{A-B}&=\int_0^1 \left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)\ln\left(\frac{2(1+x^2)}{(1+x)^2}\right)\ln x\,dx\\ &=\ln 2\int_0^1 \frac{\ln x}{1+x}\,dx-\ln 2\int_0^1 \frac{x\ln x}{1+x^2}\,dx+\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x}\,dx-\int_0^1 \frac{x\ln x\ln(1+x^2)}{1+x^2}\,dx-\\ &2\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx+ 2\int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}\,dx\\ \end{align}
En el segundo y el cuarto integrales realizar el cambio de variable $\displaystyle y=x^2$,
\begin{align} \text{A-B}&=\frac{3}{4}\ln 2\int_0^1 \frac{\ln x}{1+x}\,dx+\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x}\,dx-\frac{9}{4}\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx+ 2\int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}\,dx\\ \end{align}
Definir la función $R$ a $[0;1]$ por:
\begin{align}R(x)&=\int_0^x \frac{\ln t}{1+t}\,dt\\ &=\int_0^1 \frac{x\ln(tx)}{1+tx}\,dt\\ \end{align}
Por lo tanto,
\begin{align} \text{C}&=\int_0^1 \frac{\ln x\ln(1+x^2)}{1+x}\,dx\\ &=\Big[R(x)\ln(1+x^2)\Big]_0^1 -\int_0^1 \int_0^1 \frac{2x^2\ln(tx)}{(1+tx)(1+x^2)}\,dt\,dx\\ &=\ln 2\int_0^1 \frac{\ln x}{1+x}\,dx-\int_0^1 \int_0^1 \frac{2x^2\ln t}{(1+tx)(1+x^2)}\,dt\,dx-\int_0^1 \int_0^1 \frac{2x^2\ln x}{(1+tx)(1+x^2)}\,dt\,dx\\ &=\left(\int_0^1 \frac{2t\ln t\ln(1+t)}{1+t^2}\,dt-\int_0^1 \frac{2\ln t\ln(1+t)}{t}\,dt-\int_0^1 \frac{(\ln 2) t\ln t}{1+t^2}\,dt+\frac{\pi}{2}\int_0^1 \frac{\ln t}{1+t^2}\,dt\right)-\\ &\int_0^1 \frac{2x\ln x\ln(1+x)}{1+x^2}\,dx+\ln 2\int_0^1 \frac{\ln x}{1+x}\,dx\\ &=\ln 2\int_0^1 \frac{\ln x}{1+x}\,dx-2\int_0^1 \frac{\ln t\ln(1+t)}{t}\,dt-\ln 2\int_0^1 \frac{ t\ln t}{1+t^2}\,dt-\frac{1}{2}\pi\text{G} \end{align}
(todas las integrales han sido calculadas usando antiderivatives)
En la tercera integral realizar el cambio de variable $\displaystyle y=t^2$,
\begin{align} \text{C}&=\frac{3}{4}\ln 2\int_0^1 \frac{\ln x}{1+x}\,dx-\Big[\ln^2t\ln(1+t)\Big]_0^1+\int_0^1 \frac{\ln^2 t}{1+t}\,dt-\frac{1}{2}\pi\text{G}\\ &=\frac{3}{4}\ln 2\int_0^1 \frac{\ln x}{1+x}\,dx+\int_0^1 \frac{\ln^2 t}{1+t}\,dt-\frac{1}{2}\pi\text{G}\\ \end{align}
Definir la función $S$ a $[0;1]$ por:
\begin{align}S(x)&=\int_0^x \frac{t\ln t}{1+t^2}\,dt\\ &=\int_0^1 \frac{tx^2\ln(tx)}{1+t^2x^2}\,dt\\ \end{align}
\begin{align} \text{D}&=\int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}\,dx\\ &=\Big[S(x)\ln(1+x)\Big]_0^1-\int_0^1 \int_0^1\frac{tx^2\ln(tx)}{(1+t^2x^2)(1+x)}\,dt\,\,dx\\ &=\ln 2\int_0^1 \frac{x\ln x}{1+x^2}\,dx-\int_0^1 \int_0^1\frac{tx^2\ln t}{(1+t^2x^2)(1+x)}\,dt\,\,dx-\int_0^1 \int_0^1\frac{tx^2\ln x}{(1+t^2x^2)(1+x)}\,dt\,\,dx\\ &=\left(\int_0^1\frac{t\ln t\ln(1+t^2)}{2(1+t^2)}\,dt-\int_0^1\frac{\ln t\ln(1+t^2)}{2t}\,dt+\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt-\int_0^1\frac{(\ln 2)t\ln t}{1+t^2}\,dt\right)-\\ &\frac{1}{2}\int_0^1 \frac{\ln(1+x^2)\ln x}{1+x}\,dx+\ln 2\int_0^1 \frac{x\ln x}{1+x^2}\,dx\\ \end{align}
En la primera, segunda, cuarta, sexta integrales realizar el cambio de variable $\displaystyle y=t^2$ (o $ \displaystyle y=x^2$),
\begin{align} \text{D}&=\frac{1}{8}\int_0^1\frac{\ln t\ln(1+t)}{1+t}\,dt-\frac{1}{8}\int_0^1\frac{\ln t\ln(1+t)}{t}\,dt+\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt-\frac12 \text{C}\\ &=\frac{1}{8}\int_0^1\frac{\ln t\ln(1+t)}{1+t}\,dt-\frac{1}{16}\Big[\ln^2 t\ln(1+t)\Big]_0^1+\frac{1}{16}\int_0^1 \frac{\ln^2 t}{1+t}\,dt+\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt-\frac12 \text{C}\\ &=\frac{1}{8}\int_0^1\frac{\ln t\ln(1+t)}{1+t}\,dt+\frac{1}{16}\int_0^1 \frac{\ln^2 t}{1+t}\,dt+\int_0^1\frac{\ln t\arctan t}{1+t^2}\,dt-\frac12 \text{C}\\ \end{align}
Definir la función $T$ a $[0;1]$ por:
\begin{align}T(x)&=\int_0^x \frac{\ln t}{1+t^2}\,dt\\ &=\int_0^1 \frac{x\ln(tx)}{1+t^2x^2}\,dt\\ \end{align}
Observar que,
\begin{align}T(0)&=0\\ T(1)`&=-\text{G} \end{align}
\begin{align}\text{E}&=\int_0^1\frac{\ln x\arctan x}{1+x^2}\,dx\\ &=\Big[T(x)\arctan x\Big]_0^1-\int_0^1\int_0^1 \frac{x\ln(tx)}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\ &=-\frac{\text{G}\pi}{4}-\int_0^1\int_0^1 \frac{x\ln t}{(1+x^2)(1+t^2x^2)}\,dt\,dx-\int_0^1\int_0^1 \frac{x\ln x}{(1+x^2)(1+t^2x^2)}\,dt\,dx\\ &=-\frac{\text{G}\pi}{4}+\frac12\int_0^1 \frac{\ln t\ln\left(\frac{1+t^2}{2}\right)}{1-t^2}\,dt-E\\ \end{align}
Observe que para $t\in [0;1[$,
\begin{align}\frac{1}{1-t^2}=\frac{1}{1+t}+\frac{t}{1-t^2} \end{align}
Por lo tanto,
\begin{align}E&=-\frac{\text{G}\pi}{4}-\frac{\ln 2}{2}\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac12\int_0^1 \frac{t\ln t\ln\left(1+t^2\right)}{1-t^2}\,dt+\frac{1}{2}\text{C}-\text{E}\end{align}
En el último integral realizar el cambio de variable $\displaystyle y=t^2$
\begin{align}E&=-\frac{\text{G}\pi}{4}-\frac{\ln 2}{2}\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac18\int_0^1 \frac{\ln t\ln\left(1+t\right)}{1-t}\,dt+\frac{1}{2}\text{C}-\text{E}\\ &=-\frac{\text{G}\pi}{4}-\frac{\ln 2}{2}\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{\ln 2}{8}\int_0^1 \frac{\ln t}{1-t}\,dt+\frac18\int_0^1 \frac{\ln t\ln\left(\frac{1+t}{2}\right)}{1-t}\,dt+\\ &\frac{1}{2}\text{C}-\text{E} \end{align}
En el último integral realizar el cambio de variable $y=\dfrac{1-t}{1+t}$,
\begin{align}E&=-\frac{\text{G}\pi}{4}-\frac{\ln 2}{2}\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{\ln 2}{8}\int_0^1 \frac{\ln t}{1-t}\,dt-\\ &\frac18\int_0^1\left(\frac{1}{t}-\frac{1}{1+t}\right)\ln\left(\frac{1-t}{1+t}\right)\ln\left(1+t\right)\,dt+\frac{1}{2}\text{C}-\text{E}\\ &=-\frac{\text{G}\pi}{4}-\frac{\ln 2}{2}\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{\ln 2}{8}\int_0^1 \frac{\ln t}{1-t}\,dt+\\ &\frac18\int_0^1 \frac{\ln\left(\frac{1-t}{1+t}\right)\ln\left(1+t\right)}{1+t}\,dx+\frac18\int_0^1 \frac{\ln^2\left(1+t\right)}{t}\,dt-\\ &\frac18\int_0^1 \frac{\ln\left(1+t\right)\ln\left(1-t\right)}{t}\,dt+\frac{1}{2}\text{C}-\text{E}\\ \int_0^1 \frac{\ln^2\left(1+t\right)}{t}\,dt&=\Big[\ln t\ln^2(1+t)\Big]_0^1-2\int_0^1 \frac{\ln t\ln(1+t)}{1+t}\,dt\\ &=-2\int_0^1 \frac{\ln t\ln(1+t)}{1+t}\,dt\\ F&=\int_0^1 \frac{\ln(1+t)\ln(1-t) }{t}\,dt\\ &=\frac12 \left(\int_0^1 \frac{t\ln^2(1-t^2) }{t^2}\,dt-\int_0^1 \frac{\ln^2(1-t) }{t}\,dt-\int_0^1 \frac{\ln^2(1+t) }{t}\,dt\right)\\ \end{align}
En la primera integral realizar el cambio de variable $\displaystyle y=1-t^2$,
En la segunda integral realizar el cambio de variable $\displaystyle y=1-t$,
\begin{align}F&=\frac12 \left(2\int_0^1 \frac{\ln t\ln(1+t) }{1+t}\,dt-\int_0^1 \frac{\ln^2 t }{1-t}\,dt+\frac{1}{2}\int_0^1 \frac{\ln^2 t }{1-t}\,dt\right)\\ &=\int_0^1 \frac{\ln t\ln(1+t) }{1+t}\,dt-\frac14\int_0^1 \frac{\ln^2 t }{1-t}\,dt\\ H&=\int_0^1 \frac{\ln\left(\frac{1-t}{1+t}\right)\ln\left(1+t\right)}{1+t}\,dt \end{align}
Realizar el cambio de variable $y=\dfrac{1-t}{1+t}$,
\begin{align}H&=\int_0^1 \frac{\ln\left(\frac{2}{1+t}\right)\ln t}{1+t}\,dt\\ &=\ln 2\int_0^1 \frac{\ln t}{1+t}\,dt-\int_0^1 \frac{\ln t\ln(1+t)}{1+t}\,dt \end{align}
Por lo tanto,
\begin{align}2E&=-\frac{\text{G}\pi}{4}-\frac{\ln 2}{2}\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{\ln 2}{8}\int_0^1 \frac{\ln t}{1-t}\,dt+\\ &\frac{1}{8}\left( \ln 2\int_0^1 \frac{\ln t}{1+t}\,dt-\int_0^1 \frac{\ln t\ln(1+t)}{1+t}\,dt\right)-\frac14\int_0^1 \frac{\ln t\ln(1+t)}{1+t}\,dt-\\ &\frac18\left( \int_0^1 \frac{\ln t\ln(1+t) }{1+t}\,dt-\frac14\int_0^1 \frac{\ln^2 t }{1-t}\,dt\right)+\frac{1}{2}\text{C}\\ E&=\frac{\text{C}}{4}-\frac{\text{G}\pi}{8}-\frac{\ln 2}{8} \int_0^1\frac{\ln t}{1-t^2}\,dt+\frac{1}{64} \int_0^1\frac{\ln^2 t}{1-t}\,dt-\frac14\int_0^1 \frac{\ln t\ln(1+t)}{1+t}\,dt \\ \end{align}
Por lo tanto,
\begin{align}D&=\frac{1}{16} \int_0^1 \frac{\ln^2 t}{1+t}\,dt+\frac{1}{64} \int_0^1 \frac{\ln^2 t}{1-t}\,dt-\frac{1}{8} \int_0^1 \frac{\ln(1+t)\ln t}{1+t}\,dt-\\ &\frac{\ln 2}{8} \int_0^1 \frac{\ln t}{1-t^2}\,dt -\frac{\text{C}}{4}-\frac{\text{G}\pi} {8}\\ &=\frac{1}{64} \int_0^1 \frac{\ln^2 t}{1-t}\,dt-\frac{3}{16} \int_0^1 \frac{\ln^2 t}{1+t}\,dt-\frac{1}{8} \int_0^1 \frac{\ln(1+t)\ln t}{1+t}\,dt-\\ &\frac{\ln 2}{8} \int_0^1 \frac{\ln t}{1-t^2}\,dt-\frac{3\ln 2}{16}\int_0^1 \frac{\ln t}{1+t}\,dt\\ B&=\frac12\Big((A+B)-(A-B)\Big)\\ &=\frac{\ln^3 2}{12}+\frac{\text{G}\pi}{4}-\frac{1}{64}\int_0^1 \frac{\ln^2 t}{1-t}\,dt+\int_0^1 \frac{\ln t\ln(1+t)}{1+t}\,dt-\\ &\frac{5}{16}\int_0^1 \frac{\ln^2 t}{1+t}\,dt-\frac{5\ln 2}{16}\int_0^1 \frac{\ln t}{1+t}\,dt+\frac{\ln 2}{8}\int_0^1 \frac{\ln t}{1-t^2}\,dt\\ \end{align}
Ya que, por $t\neq 1$,
\begin{align}\frac{1}{1+t}&=\frac{1}{1-t}-\frac{2t}{1-t^2}\\ \frac{1}{1-t^2}&=\frac{1}{1-t}-\frac{t}{1-t^2} \end{align}
a continuación,
\begin{align}\text{B}&=\frac{\ln^3 2}{12}+\frac{\text{G}\pi}{4}-\frac{21}{64}\int_0^1 \frac{\ln^2 t}{1-t}\,dt+\int_0^1 \frac{\ln t\ln(1+t)}{1+t}\,dt-\\ &\frac{3\ln 2}{16}\int_0^1 \frac{\ln t}{1-t}\,dt+\frac{\ln 2}{2}\int_0^1 \frac{t\ln t}{1-t^2}\,dt+\frac{5}{8}\int_0^1 \frac{t\ln^2 t}{1-t^2}\,dt\\ \end{align}
En las dos últimas integrales realizar el cambio de variable $\displaystyle y=t^2$,
\begin{align}\text{B}&=\frac{\ln^3 2}{12}+\frac{\text{G}\pi}{4}-\frac{1}{4}\int_0^1 \frac{\ln^2 t}{1-t}\,dt+\int_0^1 \frac{\ln t\ln(1+t)}{1+t}\,dt-\\ &\frac{\ln 2}{16}\int_0^1 \frac{\ln t}{1-t}\,dt\\ U&=\int_0^1\frac{\ln(1+t)\ln t}{1+t}\,dt\\ W&=\int_0^1\frac{\ln^2\left(\frac{t}{1+t}\right)}{1+t}\,dt\\ \end{align}
Realizar el cambio de variable $y=\dfrac{t}{1+t}$,
\begin{align} W&=\int_0^{\frac{1}{2}}\frac{\ln^2 t}{1-t}\,dt\\ &=\int_0^1\frac{\ln^2 x}{1-t}\,dt-\int_{\frac{1}{2}}^1\frac{\ln^2 t}{1-t}\,dt\\ \end{align}
En el último integral realizar el cambio de variable $y=\dfrac{1-t}{t}$
\begin{align} W&=\int_0^1\frac{\ln^2 t}{1-t}\,dt-\int_0^1\frac{\ln^2(1+t)}{t(1+t)}\,dt\\ &=\int_0^1\frac{\ln^2 t}{1-t}\,dt+\int_0^1\frac{\ln^2(1+t)}{1+t}\,dt-\int_0^1\frac{\ln^2(1+t)}{t}\,dt\\ &=\int_0^1\frac{\ln^2 t}{1-t}\,dt+\int_0^1\frac{\ln^2(1+t)}{1+t}\,dt-\Big[\ln t\ln^2(1+t)\Big]_0^1+2\int_0^1 \frac{\ln x\ln(1+t)}{1+t}\,dt\\ &=\int_0^1\frac{\ln^2 t}{1-t}\,dt+\int_0^1\frac{\ln^2(1+t)}{1+t}\,dt+2U\\ \end{align}
Por otro lado,
\begin{align} W&=\int_0^1\frac{\left(\ln t-\ln(1+t)\right)^2}{1+t}\,dt\\ &=\int_0^1\frac{\ln^2 t}{1+t}\,dt+\int_0^1\frac{\ln^2(1+t)}{1+t}\,dx-2U\\ \end{align}
Por lo tanto,
\begin{align} U&=\frac{1}{4}\left(\int_0^1\frac{\ln^2 t}{1+t}\,dt-\int_0^1\frac{\ln^2 t}{1-t}\,dt\right)\\ &=-\frac{1}{4}\int_0^1\frac{2t\ln^2 t}{1-t^2}\,dt \end{align}
Realizar el cambio de variable $\displaystyle y=t^2$,
\begin{align} U&=-\frac{1}{16}\int_0^1\frac{\ln^2 t}{1-t}\,dt\\ \end{align}
Por lo tanto,
\begin{align}\text{B}&=\frac{\ln^3 2}{12}+\frac{\text{G}\pi}{4}-\frac{5}{16}\int_0^1 \frac{\ln^2 t}{1-t}\,dt-\frac{\ln 2}{16}\int_0^1 \frac{\ln t}{1-t}\,dt\\ &=\frac{\ln^3 2}{12}+\frac{\text{G}\pi}{4}-\frac{5}{16}\times 2\zeta(3)-\frac{\ln 2}{16}\times -\frac{\pi^2}{6}\\ &=\boxed{\frac{1}{12}\ln^3 2+\frac{1}{4}\text{G}\pi-\frac{5}{8}\zeta(3)+\frac{1}{96}\pi^2\ln 2} \end{align}
Espero que no te importe mi intento de la integral indefinida mediante polylogarithms. NO UNA RESPUESTA
Me re-escribió $\log^2(1+x^2)$como $$\log^2(1+ix)(1-ix)=\log^2(1+ix) + 2\log(1+ix)\log(1-ix) + \log^2(1-ix).$$ Por lo tanto, se puede escribir \begin{eqnarray} \mathcal I &=& \int\frac{\log^2(1+x^2)}{1+x}dx=\\ &=& \underbrace{\int\frac{\log^2(1+ix)}{1+x}dx}_{\mathcal I_1} + 2\underbrace{\int\frac{\log(1+ix)\log(1-ix)}{1+x}dx}_{\mathcal I_2}+\\ & &+\underbrace{\int\frac{\log^2(1-ix)}{1+x}dx}_{\mathcal I_3}\tag{*}\label{3} \end{eqnarray} Yo se concentró primero en la integral $$\mathcal I_1 = \int \frac{\log^2(1+ix)}{1+x}dx$$ para el que he utilizado el cambio de variables $$\omega = \frac{x+1}{2} -i \frac{x+1}{2},$$ que da $$ x = (1+i)\omega -1,$$ $$1+ix = (1-i)(\omega + 1),$$ y $$dx = (1+i)d\omega.$$ Así, obtenemos \begin{eqnarray} \mathcal I_1 &=& \int \frac{\left[\log(1-i)+\log(1+\omega)\right]^2}{\omega}d\omega=\\ &=&\log^2(1-i)\log \omega +2\log(1-i) \text{Li}_2(-\omega) +\underbrace{\int\frac{\text{Li}_1^2(-\omega)}{\omega}d\omega}_{\mathcal A(\omega)}. \end{eqnarray} La integración por partes $\mathcal A$rendimientos \begin{eqnarray} \mathcal A(\omega) &=& \text{Li}_1^2(-\omega)\log(-\omega) + 2\int\frac{\text{Li}_1(-\omega)\log(-\omega)}{1+\omega}d\omega=\\ &=&\text{Li}_1^2(-\omega)\log(-\omega) + 2\int \text{Li}_1(-\omega)d\text{Li}_2(1+\omega)=\\ &=&\text{Li}_1^2(-\omega)\log(-\omega)+2\text{Li}_1(-\omega)\text{Li}_2(1+\omega) + 2\int \frac{\text{Li}_2(1+\omega)}{1+\omega}d\omega=\\ &=&\text{Li}_1^2(-\omega)\log(-\omega)+2\text{Li}_1(-\omega)\text{Li}_2(1+\omega) + \text{Li}_3(1+\omega). \end{eqnarray} Así tenemos \begin{eqnarray} \mathcal I_1 &=&\log^2(1-i)\log \omega +2\log(1-i) \text{Li}_2(-\omega) +\\ &&+\text{Li}_1^2(-\omega)\log(-\omega)+2\text{Li}_1(-\omega)\text{Li}_2(1+\omega) + \text{Li}_3(1+\omega) \end{eqnarray}
Ahora, utilizando el mismo cambio de variable en $\mathcal I_3$ tenemos \begin{eqnarray} \mathcal I_3 &=& \int \frac{\left[\log(1+i)+\log(1-i\omega)\right]^2}{\omega}d\omega=\\ &=&\log^2(1+i)\log \omega +2\log(1+i) \text{Li}_2(i\omega) +\underbrace{\int\frac{\text{Li}_1^2(i\omega)}{\omega}d\omega}_{\mathcal B(\omega)}. \end{eqnarray} Observar que $$\mathcal B(\omega) = i\mathcal A(-i\omega).$$ Por lo tanto, podemos utilizar los resultados ya obtenidos para obtener \begin{eqnarray} \mathcal I_3 &=& \log^2(1+i)\log \omega +2\log(1+i) \text{Li}_2(i\omega) +\\ &&+i\text{Li}_1^2(i\omega)\log(i\omega)+2i\text{Li}_1(i\omega)\text{Li}_2(1-i\omega) + i\text{Li}_3(1-i\omega) \end{eqnarray}
De nuevo el mismo cambio en las variables de rendimiento, para $\mathcal I_2$, \begin{eqnarray} \mathcal I_2 &=& \int\frac{[\log(1-i)+\log(1+\omega)][\log(1+i)+\log(1-i\omega)}{\omega}d\omega=\\ &=& \log 2 \log \omega + \log(i-1)\text{Li}_2(i\omega)+\\ &&+\log(i+1)\text{Li}_2(-\omega)+\underbrace{\int\frac{\log(1-i\omega)\log(1+\omega)}{\omega}d\omega}_{\mathcal C}. \end{eqnarray} $\mathcal C$ es la parte más delicada. No he terminado todavía.
