Encuentra esta integral $\int_{\frac{1}{2}}^{1} \dfrac{x\ln x}{1+x^2} dx$

Question

Necesito encontrar el valor de : $$\int_{\frac{1}{2}}^{1} \dfrac{x\ln x}{1+x^2} dx$$ He intentado la integración por partes pero necesito encontrar $$\int_{\frac{1}{2}}^{1} \dfrac{\ln(1+x^2)}{x} dx$$

¿Tiene alguna idea?

Answer 1

Hacer el cambio de variable $x^2=y$ tenemos $$\begin{align*} \int^1_{\frac 12} \frac{\ln(1+x^2)}{x} \mathrm dx&=\frac12\int^1_{\frac 14}\frac{\ln(1+y)}{y}\mathrm dy \\&=\frac 12 \int^1_{\frac 14}\text{Li}_2'(-y)\mathrm dy \\&=\frac 12\left[-\text{Li}_2(-y)\right]^1_{\frac 14} \\&=\frac12\left(\text{Li}_2\left(-\frac 14\right)-\text{Li}_2(-1)\right) \\&=\frac12\left(\text{Li}_2\left(-\frac 14\right)+\frac{\pi^2}{12}\right), \end{align*}$$ donde $ \text{Li}_2(z)=\sum_{n=1}^\infty \frac{z^n}{n^2},\ |z|\le 1 $ y $\text{Li}_2(-1)=\sum_{n=1}^\infty \frac{(-1)^n}{n^2}=-\frac12\zeta(2)=-\frac{\pi^2}{12}.$

Así, la integral dada es $$\begin{align*} \int_{\frac{1}{2}}^{1} \dfrac{x\ln x}{1+x^2} \mathrm dx&=\left[\frac{\ln(1+x^2)}2\ln x\right]^1_{\frac12}-\frac12\int^1_{\frac12} \frac{\ln(1+x^2)}{x} \mathrm dx \\&=\frac{\ln\left(\frac54\right)\ln 2}{2}-\frac 14\left(\text{Li}_2\left(-\frac 14\right)+\frac{\pi^2}{12}\right). \end{align*}$$

Answer 2

\begin{align}J&=\int_{\frac{1}{2}}^1\frac{\ln(1+x^2)}{x}\,dx\end{align}

Realizar el cambio de variable $u=x^2$ ,

\begin{align}J&=\frac{1}{2}\int_{\frac{1}{4}}^1\frac{\ln(1+u)}{u}\,du\\ &=\frac{1}{2}\Big[\ln(1+u)\ln u\Big]_{\frac{1}{4}}^1-\frac{1}{2}\int_{\frac{1}{4}}^1\frac{\ln u}{1+u}\,du\\ &=\ln\left(\frac{5}{4}\right)\ln 2-\frac{1}{2}\int_0^1\frac{\ln u}{1+u}\,du+\frac{1}{2}\int_0^{\frac{1}{4}}\frac{\ln u}{1+u}\,du \\\end{align}

En esta última integral realizar el cambio de variable $t=4u$ ,

\begin{align}J&=\ln\left(\frac{5}{4}\right)\ln 2-\frac{1}{2}\int_0^1\frac{\ln u}{1+u}\,du+\frac{1}{8}\int_0^{1}\frac{\ln\left(\frac{1}{4}v\right)}{1+\frac{1}{4}v}\,dv\\ &=\ln\left(\frac{5}{4}\right)\ln 2-\frac{1}{2}\int_0^1\frac{\ln u}{1+u}\,du-\ln 2\Big[\ln\left(1+\frac{1}{4}v\right)\Big]_0^1+\frac{1}{8}\int_0^1\frac{\ln v}{1+\frac{1}{4}v}\,dv\\ &=\frac{1}{8}\int_0^1\frac{\ln v}{1+\frac{1}{4}v}\,dv-\frac{1}{2}\int_0^1\frac{\ln u}{1+u}\,du\\ &=\frac{1}{8}\int_0^1\left(\frac{1}{1-\frac{1}{4}v}-\frac{v}{2\left(1-\frac{1}{16}v^2\right)}\right)\ln v\,dv-\frac{1}{2}\int_0^1\left(\frac{1}{1-u}-\frac{2u}{1-u^2}\right)\ln u\,du\\ &=\frac{1}{8}\int_0^1 \frac{\ln v}{1-\frac{1}{4}v}\,dv-\frac{1}{2}\int_0^1 \frac{\ln v}{1-v}\,dv-\frac{1}{16}\int_0^1\frac{v\ln v}{1-\frac{1}{16}v^2}\,dv+\int_0^1 \frac{v\ln v}{1-v^2} \,dv\\ \end{align}

En las dos últimas integrales realizar el cambio de variable $y=v^2$ ,

\begin{align}J&=\frac{1}{8}\int_0^1 \frac{\ln y}{1-\frac{1}{4}y}\,dy-\frac{1}{2}\int_0^1 \frac{\ln v}{1-v}\,dv-\frac{1}{64}\int_0^1\frac{\ln y}{1-\frac{1}{16}y}\,dy+\frac{1}{4}\int_0^1 \frac{\ln y}{1-y} \,dy\\ &=\frac{1}{8}\int_0^1 \frac{\ln y}{1-\frac{1}{4}y}\,dy-\frac{1}{4}\int_0^1 \frac{\ln v}{1-v}\,dv-\frac{1}{64}\int_0^1\frac{\ln y}{1-\frac{1}{16}y}\,dy\\ &=\frac{1}{4}\text{Li}_2\left(\frac{1}{16}\right)+\frac{1}{4}\text{Li}_2\left(1\right)-\frac{1}{2}\text{Li}_2\left(\frac{1}{4}\right)\\ &=\boxed{\frac{\pi^2}{24}-\frac{1}{2}\text{Li}_2\left(\frac{1}{4}\right)+\frac{1}{4}\text{Li}_2\left(\frac{1}{16}\right)} \end{align}

NB:

Lo asumo,

\begin{align}\text{Li}_2\left(1\right)=\frac{\pi^2}{6}\\ \int_0^1 \frac{\ln^r t}{1-at}\,dt&=\frac{(-1)^r r!}{a}\text{Li}_{r+1}\left(a\right) \end{align}

$r\geq 1$ entero, $|a|\leq 1$ complejo, $a\neq 0$ .

Para una definición de $\text{Li}_2\left(x\right)$ ver: https://en.wikipedia.org/wiki/Polylogarithm

Si prefiere una respuesta más corta,

\begin{align}\boxed{J=\frac{1}{2}\text{Li}_2\left(-\frac{1}{4}\right)+\frac{\pi^2}{24}}\end{align}