Demasiado tiempo para comentar.
\begin{align*}
I=\int_0^\infty \frac{x \ln(1+x)}{(1+x)(2x^2+2x+1)} dx=\frac{5\pi^2}{96}-\frac{\ln^2 2} 8.
\end{align*}
Aquí están algunas de las formas alternativas a evaluar. (Yo prefiero el método 2)
Método 1: Serie + Residuo
Sustituto $x=\frac 1t$,
\begin{align*}
I &= \int_0^\infty \frac{\ln(1+t)}{(1+t)(t^2+2t+2)} dt-\int_0^\infty \frac{\ln t}{(1+t)(t^2+2t+2)} dt \\
&= I_1-I_2.
\end{align*}
Reivindicación 1:
\begin{align*}
I_1=\int_0^\infty \frac{\ln(1+t)}{(1+t)(t^2+2t+2)} dt=\frac{\pi^2}{48}.
\end{align*}
Poner $t=-1+\frac 1u$$I_1$,
\begin{align*}
I_1 &= \int_0^1 -\frac{u\ln u}{1+u^2}du=- \int_0^1 \sum_{k=0}^\infty (-1)^k u^{2k+1}\ln u\,du\\
&=- \sum_{k=0}^\infty (-1)^k \left[\frac{u^{2k+2}\ln u}{2k+2}-\frac{u^{2k+2}}{(2k+2)^2}\right]_0^1\\
&=\frac 14 \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)^2} \tag {1.1}\\
&=\frac 14\cdot\frac{\pi^2}{12}=\frac{\pi^2}{48}.
\end{align*}
Comentario:
Reivindicación 2:
\begin{align*}
I_2=\int_0^\infty \frac{\ln t}{(1+t)(t^2+2t+2)} dt=\frac{\ln^2 2}{8}-\frac{\pi^2}{32}.
\end{align*}
Considerar la llave de agujero de contorno como se muestra:
![enter image description here]()
Debido a $\frac{\partial}{\partial\alpha}t^\alpha=t^\alpha\ln t$, se utiliza la función
\begin{align*}
f(z)=\frac{z^\alpha}{(1+z)(z^2+2z+2)},\qquad\forall -1<\alpha<1.
\end{align*}
Los polos de $f$$-1$$-1\pm i$.
Conjunto
\begin{align*}
I_2(\alpha) =\int_0^\infty f(t)\,dt=\int_0^\infty \frac{t^\alpha}{(1+t)(t^2+2t+2)}\,dt.
\end{align*}
A continuación,
\begin{align*}
2\pi i\sum \text{Res } f&=\oint f(z) dz\\
&=\int_\gamma+\int_\epsilon^R\frac{z^\alpha}{(1+z)(z^2+2z+2)}\,dz+\int_\Gamma+\int_R^\epsilon\frac{\left(ze^{2\pi i}\right)^\alpha}{(1+z)(z^2+2z+2)}\,dz\\
&=\left(1-e^{2\pi\alpha i}\right)\int_\epsilon^R\frac{z^\alpha}{(1+z)(z^2+2z+2)}\,dz+\left(\int_\gamma+\int_\Gamma\right)\\
&=\left(1-e^{2\pi\alpha i}\right)I_2(\alpha).
\end{align*}
Y,
\begin{align*}
I_2&=\left.\frac{\partial}{\partial \alpha}\right|_{\alpha=0}I_2(\alpha)\\
&=\left.\frac{\partial}{\partial \alpha}\right|_{\alpha=0}\frac{\pi\left(2^{\frac {\alpha}2}\cos\left(\frac{\pi \alpha}4\right)-1\right)}{\sin(\pi \alpha)}\\
&=\frac{\ln^2 2}{8}-\frac{\pi^2}{32}.
\end{align*}
O
Uso el mismo contorno y considerar la función $g$:
\begin{align*}
g(z)=\frac{\ln^2 z}{(1+z)(z^2+2z+2)}.
\end{align*}
Los polos de $g$$-1$$-1\pm i$.
El uso de $0\le\arg z<2\pi$ como la rama del logaritmo correspondiente.
\begin{align*}
2\pi i\sum \text{Res } g&=\oint g(z) dz\\
&=\int_\gamma+\int_\epsilon^R\frac{(\ln z)^2}{(1+z)(z^2+2z+2)}\,dz+\int_\Gamma+\int_R^\epsilon\frac{(\ln z+2\pi i)^2}{(1+z)(z^2+2z+2)}\,dz\\
&=\int_\epsilon^R\frac{(\ln z)^2-(\ln z+2\pi i)^2}{(1+z)(z^2+2z+2)}\,dz+\left(\int_\gamma+\int_\Gamma\right)\\
&=-4\pi i\int_0^\infty\frac{\ln z}{(1+z)(z^2+2z+2)}\,dz+\int_0^\infty\frac{4\pi^2}{(1+z)(z^2+2z+2)}\,dz\\
&=-4\pi iI_2+\int_0^\infty\frac{4\pi^2}{(1+z)(z^2+2z+2)}\,dz.
\end{align*}
Método 2: Diferenciación en $\int$
Sustituto $x=\frac{u}{1-u}$,
\begin{align*}
I&=-\int_0^1 \frac{u\ln(1-u)}{u^2+1}\,du.
\end{align*}
Escribir
\begin{align*}
J(\beta)&=-\int_0^1 \frac{u\ln(1-\beta u)}{u^2+1}\,du,\qquad \forall\beta\in[0,1].
\end{align*}
A continuación, $J(0)=0$ y
\begin{align*}
J_\beta (\beta)&=\frac{\partial}{\partial \beta}J(\beta)=\int_0^1 \frac{u^2}{(1-\beta u)\left(u^2+1\right)}\,du\\
&=\frac 1{1+\beta^2}\left(\color{red}{-\int_0^1\frac{1+\beta u}{1+u^2}\,du}\color{green}{+\int_0^1 \frac{1}{1-\beta u}\,du}\right)\\
&=\color{red}{-\frac{\pi+2\beta \ln 2}{4(1+\beta^2)}}\color{green}{-\frac{\ln(1-\beta)}{\beta}+\frac{\beta\ln(1-\beta)}{1+\beta^2}}.
\end{align*}
\begin{align*}
I&=J(1)=\int_0^1 J_\beta (\beta)\,d\beta\\
&=\color{red}{-\int_0^1\frac{\pi+2\beta \ln 2}{4(1+\beta^2)}\,d\beta}\color{blue}{-\int_0^1\frac{\ln(1-\beta)}{\beta}\,d\beta}+\int_0^1\frac{\beta\ln(1-\beta)}{1+\beta^2}\,d\beta\\
&=\color{red}{-\frac{\pi^2}{16}-\frac 14\ln^2 2}\color{blue}{+\frac{\pi^2}{6}}-I.
\end{align*}
\begin{align*}
\therefore 2I=\frac{5\pi^2}{48}-\frac 14\ln^2 2.
\end{align*}
Método 3 : Serie
Inspirado por $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$.
Reivindicación 3:
\begin{align*}
I=-\frac 12 \sum_{n=1}^\infty\frac{(-1)^n} n H_{2n},\qquad\qquad\text{where }H_m=\sum_{n=1}^m\frac 1 n,\quad m\in\mathbb N.
\end{align*}
Sustituto $x=\frac{u}{1-u}$,
\begin{align*}
I&=-\int_0^1 \frac{u\ln(1-u)}{u^2+1}\,du.
\end{align*}
Integrar por partes,
\begin{align*}
I&=-\frac 12\int_0^1 \ln(1-u)\,d\left(\ln \frac {1+u^2}{2}\right)\\
&=\underbrace{\left.-\frac 12\ln(1-u)\left(\ln \frac {1+u^2}{2}\right)\right|_0^1}_0+\frac 12 \int_0^1\frac{\color{red}{\ln\left(1+u^2\right)}-\color{green}{\ln 2}}{u-1}\,du\\
&=\frac 12 \int_0^1 \frac 1{u-1}\left(\color{red}{\sum_{n=1}^\infty\frac{(-1)^{n-1}} n u^{2n}}-\color{green}{\sum_{n=1}^\infty\frac{(-1)^{n-1}} n }\right)\,du\\
&=-\frac 12 \sum_{n=1}^\infty\frac{(-1)^n} n\int_0^1\frac {u^{2n}-1}{u-1}\,du\\
&=-\frac 12 \sum_{n=1}^\infty\frac{(-1)^n} n\int_0^1\sum_{k=1}^{2n} u^{k-1}\,du\\
&=-\frac 12 \sum_{n=1}^\infty\frac{(-1)^n} n H_{2n}.
\end{align*}
La reclamación se ha establecido.
Deje $S(N)=\sum_{n=1}^N\frac{(-1)^n} n H_{2n},\forall N\in\mathbb N$.
\begin{align*}
S(N)&=\sum_{n=1}^N\frac{(-1)^n} n \cdot \frac 1{2n}+\sum_{n=1}^N\frac{(-1)^n} n H_{2n-1}\tag{3.1}\\
&=\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}+\frac 12\sum_{n=1}^N\sum_{k=1}^{2n-1}\frac{(-1)^n} n\left(\frac 1k+\frac 1{2n-k}\right)\tag{3.2}\\
&=\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}+\sum_{n=1}^N\sum_{k=1}^{2n-1}\frac{(-1)^n}{k(2n-k)}\tag{3.3}\\
&=\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}+\Re\left\{\sum_{k=1}^{2N-1}\sum_{m=k+1}^{2N}\frac{(-1)^{m/2}}{k(m-k)}\right\}\tag{3.4}\\
&=\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}+\Re\left\{\sum_{k=1}^{2N-1}\sum_{l=1}^{2N-k}\frac{i^{k+l}}{kl}\right\}\tag{3.5}\\
&=\color{purple}{\frac 12\sum_{n=1}^N\frac{(-1)^n}{n^2}}+\color{blue}{\Re\left\{\sum_{k=1}^{2N-1}\frac{i^k}k\sum_{l=1}^{2N-1}\frac{i^l}l\right\}}-\color{orange}{\Re\left\{\sum_{k=1}^{2N-1}\frac{i^k}k\sum_{l=2N-k-1}^{2N-1}\frac{i^l}l\right\}}\tag{3.6}\\
&=\color{purple}{S_1(N)}+\color{blue}{S_2(N)}-\color{orange}{S_3(N)}.
\end{align*}
Comentario:
En $(3.1)$ rompemos $H_{2n}$ a $\frac 1{2n}+H_{2n-1}$.
En $(3.2)$$\sum_{k=1}^{m-1}f(k)=\sum_{k=1}^{m-1}f(m-k)$.
En $(3.4)$ intercambio de dos signos de suma y conjunto de $m=2n$.
En $(3.5)$ establecer $l=m-k$.
En $(3.6)$$\sum_{l=1}^{2N-k}=\sum_{l=1}^{2N-1}-\sum_{l=2N-k-1}^{2N-1}$.
Para demostrar $\color{orange}{S_3(\infty)}=0$, el uso de la Alternancia de Serie de la Prueba, uno tiene
\begin{align*}
\left|\color{orange}{S_3(N)}\right|=&\left|\Re\left\{\sum_{k=1}^{2N-1}\left(\frac{i^{2k}}{2k}+\frac{i^{2k-1}}{2k-1}\right)\sum_{l=2N-k-1}^{2N-1}\left(\frac{i^{2l}}{2l}+\frac{i^{2l-1}}{2l-1}\right)\right\}\right|\tag{3.7}\\
&=\left|\Re\left\{\sum_{k=1}^{2N-1}\left(\frac{(-1)^k}{2k}-i\frac{(-1)^k}{2k-1}\right)\sum_{l=2N-k-1}^{2N-1}\left(\frac{(-1)^l}{2l}-i\frac{(-1)^l}{2l-1}\right)\right\}\right|\tag{3.8}\\
&=\left|\sum_{k=1}^{2N-1}\frac{(-1)^k}{2k}\sum_{l=2N-k+1}^{2N-1}\frac{(-1)^l}{2l}+\sum_{k=1}^{2N-1}\frac{(-1)^k}{2k-1}\sum_{l=2N-k-1}^{2N-1}\frac{(-1)^l}{2l-1}\right|\tag{3.9}\\
&\le\left|\sum_{k=1}^{2N-1}\frac{(-1)^k}{k}\sum_{l=2N-k-1}^{2N-1}\frac{(-1)^l}{l}\right|\le\left|\sum_{k=1}^{2N-1}\frac{(-1)^k}{k}\sum_{l=N}^{2N-1}\frac{(-1)^l}{l}\right|\\
&\le\left|2\sum_{l=N}^{2N-1}\frac1{N(N-1)}\right|= \frac 2{N-1}.
\end{align*}
Comentario:
- En $(3.7)$ rompemos índices en pares e impares de las partes.
Por lo tanto, como $N\to\infty$,
\begin{align*}
\lim_{N\to\infty} S(N)&=\color{purple}{S_1(\infty)}+\color{blue}{S_2(\infty)}-\color{orange}{S_3(\infty)}\\
&=\color{purple}{\frac 12\left(-\frac{\pi^2}{12}\right)}+\color{blue}{\Re\left\{\ln(1-i)\ln(1-i)\right\}}-\color{orange}{0}\\
&=\color{purple}{-\frac{\pi^2}{24}}+\color{blue}{\left(\frac{\ln 2}{2}\right)^2+\left(\frac{-\pi i}{4}\right)^2}-\color{orange}{0}\\
&=-\frac{5\pi^2}{48}+\frac{\ln^2 2} 4.
\end{align*}
