Integración de$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx$ con técnicas restringidas

Question

Nuestro profesor nos pide que calculemos esta integral: $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx$$ Using only changes of variables, integrations by parts and the following result: $$ \int_0^1 \frac{\ln x}{x-1}\,dx=\frac{\pi^2}{6}$ $ (Sin usar análisis complejo, series, derivado bajo el signo integral, integrales dobles o funciones especiales).

Algunos métodos que no respetan el requisito del maestro se encuentran aquí .

Answer 1

Considere las siguientes integrales:$$I=\int_0^1\frac{x\ln (1+x)}{1+x^2}dx,\quad J=\int_0^1\frac{x\ln (1-x)}{1+x^2}dx$$

---

$$I+J=\int_0^1 \frac{x\ln(1-x^2)}{1+x^2}dx\overset{x^2=t}=\frac12\int_0^1 \frac{\ln(1-t)}{1+t}dt$$ Ahora vamos a integral por partes, sin embargo, no podemos eligió $\ln(1+t)'=\frac{1}{1+t}$ ya que nos encontramos con problemas de divergencia.

Tomaremos $(\ln (1+t)-\ln 2)'=\frac{1}{1+t}$luego: $$I+J=\frac12\bigg[\underbrace{\ln(1-t)(\ln(1+t)-\ln 2)\bigg]_0^1}_{=0}+\frac12 \int_0^1 \frac{\ln\left(\frac{1+t}{2}\right)}{1-t}dt$$ Ahora sustituye $t=\frac{1-y}{1+y}$ para conseguir: $$I+J=-\frac12 \int_0^1 \frac{\ln(1+y)}{y(1+y)}dy=\frac12 \int_0^1 \frac{\ln(1+y)}{1+y}dy-\frac12\int_0^1 \frac{\ln(1+y)}{y}dy$$ Se sabe a partir de aquíque: $$\int_0^1 \frac{\ln(1+x)}{x}dx=\frac12 \int_0^1 \frac{\ln x}{x-1}dx=\frac12\cdot \frac{\pi^2}{6}$$ $$\Rightarrow I+J=\frac14 \ln^2(1+y)\bigg|_0^1 -\frac14 \cdot \frac{\pi^2}{6}=\boxed{\frac{\ln^2 2}{4}-\frac{\pi^2}{24}}$$

---

Del mismo modo, para $I-J$ establecer $\frac{1-x}{1+x}= t$ para obtener: $$I-J=-\int_0^1 \frac{x\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}dx=\underbrace{\int_0^1 \frac{t\ln t}{1+t^2}dt}_{t^2\rightarrow t}-\int_0^1 \frac{\ln t}{1+t}dt$$ $$=\frac14 \int_0^1 \frac{\ln t}{1+t}dt-\int_0^1 \frac{\ln t}{1+t}dt=-\frac34 \left(\underbrace{\ln(1+t)\ln t \bigg|_0^1}_{=0} -\int_0^1 \frac{\ln(1+t)}{t}dt\right)$$ $$=\frac34 \int_0^1 \frac{\ln(1+t)}{t}dt=\frac38 \int_0^1 \frac{\ln t}{t-1}dt=\boxed{\frac{\pi^2}{16}}$$

---

Finalmente podemos extraer la integral como: $$I=\frac12 \left((I+J)+(I-J)\right)=\frac12\left(\frac{\ln^2 2}{4}-\frac{\pi^2}{24}+\frac{\pi^2}{16}\right)=\boxed{\frac{\ln^2 2}{8}+\frac{\pi^2}{96}}$$ Complementaria resultado es el siguiente: $$\boxed{\int_0^1\frac{x\ln (1-x)}{1+x^2}dx=\frac{\ln^2 2}{8}-\frac{5\pi^2}{96}}$$

Answer 2

Solución alternativa:

\begin{align} J&=\int_0^1 \frac{x\ln(1+x)}{1+x^2}\,dx\\ A&=\int_0^1 \frac{\ln(1+x^2)}{1+x}\,dx\\ &=\Big[\ln(1+x^2)\ln(1+x)\Big]_0^1-2\int_0^1 \frac{x\ln(1+x)}{1+x^2}\,dx\\ &=\ln^2 2-2J\\ J&=\frac{1}{2}\ln^2 2-\frac{1}{2}A\\ B&=\int_0^1 \frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{1+x}\,dx\\ C&=\int_0^1 \frac{\ln\left(1-x\right)}{1+x}\,dx\\ \end{align} Realizar el cambio de variable $y=\dfrac{1-x^2}{1+x^2}$, \begin{align} B&=\frac{1}{2}\int_0^1 \frac{\left(\sqrt{1+x}-\sqrt{1-x}\right)\ln x}{x\sqrt{1-x}(1+x)}\,dx\\ &=\frac{1}{2}\int_0^1 \left(\frac{1}{\sqrt{1-x^2}}-\frac{1}{1+x}\right)\frac{\ln x}{x}\,dx\\ &=\frac{1}{2}\Big[\left(\ln(1+x)-\ln(1+\sqrt{1-x^2})+\ln 2\right)\ln x\Big]_0^1-\\ &\frac{1}{2}\int_0^1\frac{\ln(1+x)-\ln(1+\sqrt{1-x^2})+\ln 2}{x}\,dx\\ &=-\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x}\,dx+\frac{1}{2}\int_0^1\left(\frac{\ln(1+\sqrt{1-x^2})}{x}-\frac{\ln 2}{x}\right)\,dx \end{align} En la segunda integral, realizar el cambio de variable $y=\dfrac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}$, \begin{align} B&=-\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x}\,dx-\frac{1}{4}\int_0^1 \frac{(1-x)\ln(1+x)}{x(1+x)}\,dx\\ &=-\frac{1}{2}\int_0^1 \frac{\ln(1+x)}{x}\,dx-\frac{1}{4}\int_0^1\left(\frac{\ln(1+x)}{x}-\frac{2\ln(1+x)}{1+x}\right)\,dx\\ &=-\frac{3}{4}\int_0^1 \frac{\ln(1+x)}{x}\,dx+\frac{1}{4}\Big[\ln^2(1+x)\Big]_0^1\\ &=\left(-\frac{3}{4}\Big[\ln x\ln(1+x)\Big]_0^1+\frac{3}{4}\int_0^1 \frac{\ln x}{1+x}\,dx\right)+\frac{1}{4}\ln^2 2\\ &=\frac{3}{4}\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{1}{4}\ln^2 2\\ \end{align} Realizar el cambio de variable $y=\dfrac{1-x}{1+x}$, \begin{align} C&=\int_0^1 \frac{\ln\left(\frac{2x}{1+x}\right)}{1+x}\,dx\\ &=\int_0^1 \frac{\ln 2}{1+x}\,dx+\int_0^1 \frac{\ln x}{1+x}\,dx-\int_0^1 \frac{\ln(1+x)}{1+x}\,dx\\ &=\ln^2 2+\int_0^1 \frac{\ln x}{1+x}\,dx-\frac{1}{2}\ln^2 2\\ &=\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{1}{2}\ln^2 2\\ B&=\int_0^1 \frac{\ln\left(\frac{(1-x)(1+x)}{1+x^2}\right)}{1+x}\,dx\\ &=C+\int_0^1 \frac{\ln(1+x)}{1+x}\,dx-A\\ &=C+\frac{1}{2}\ln^2 2-A \end{align} Por lo tanto, \begin{align}A&=C+\frac{1}{2}\ln^2 2-B\\ &=\left(\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{1}{2}\ln^2 2\right)+\frac{1}{2}\ln^2 2-\left(\frac{3}{4}\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{1}{4}\ln^2 2\right)\\ &=\frac{1}{4}\int_0^1 \frac{\ln x}{1+x}\,dx+\frac{3}{4}\ln^2 2\\ &=\frac{1}{4}\left(\int_0^1 \frac{\ln x}{1-x}\,dx-\frac{2x\ln x}{1-x^2}\,dx\right)+\frac{3}{4}\ln^2 2 \end{align} En la segunda integral realizar el cambio de variable $y=x^2$, \begin{align}A&=\frac{1}{4}\left(\int_0^1 \frac{\ln x}{1-x}\,dx-\frac{1}{2}\int_0^1\frac{\ln x}{1-x}\,dx\right)+\frac{3}{4}\ln^2 2\\ &=\frac{1}{8}\int_0^1\frac{\ln x}{1-x}\,dx+\frac{3}{4}\ln^2 2\\ J&=\frac{1}{2}\ln^2 2-\frac{1}{2}\left(\frac{1}{8}\int_0^1\frac{\ln x}{1-x}\,dx+\frac{3}{4}\ln^2 2\right)\\ &=\frac{1}{8}\ln^2 2-\frac{1}{16}\int_0^1\frac{\ln x}{1-x}\,dx\\ &=\boxed{\frac{1}{8}\ln^2 2+\frac{1}{96}\pi^2}\\ \end{align}