Necesita ayuda con $\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx$

Question

Yo necesito ayudar con esta integral: $$\int_0^\infty\frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}dx.$$ Mathematica dice que no convergen, lo cual es falso.

Answer 1

Enorme dolor en la rabadilla, pero en el final, bastante sencillo. Primero observar que

$$\begin{align}\int_0^{\infty} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} &= \int_0^{1} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} + \int_1^{\infty} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)}\\ &= \int_0^{1} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} + \int_0^{1} dx \, x^3 \frac{\log{(1+x)}-\log{x}}{(1+x^2)(1+x^3)}\\ &= \int_0^1 dx \frac{\log{(1+x)}}{1+x^2} - \int_0^1 dx \frac{x^3 \log{x}}{(1+x^2)(1+x^3)}\end{align} $$

La primera integral se puede evaluar por subbing $x=\tan{t}$:

$$\begin{align} \int_0^1 dx \frac{\log{(1+x)}}{1+x^2} &= \int_0^{\pi/4} dt \, \log{(1+\tan{t})}\\ &= \int_0^{\pi/4} dt \, \log{(\sin{t}+\cos{t})} - \int_0^{\pi/4} dt \, \log{(\cos{t})}\\ &= \int_0^{\pi/4} dt \, \log{(\sqrt{2} \cos{(t-\pi/4)})} - \int_0^{\pi/4} dt \, \log{(\cos{t})}\\ &= \frac{\pi}{8} \log{2} + \int_0^{\pi/4} dt \, \log{(\cos{(t-\pi/4)})} - \int_0^{\pi/4} dt \, \log{(\cos{t})}\\ &= \frac{\pi}{8} \log{2}\end{align}$$

La segunda integral es complicado pero se puede hacer. En primer lugar, podemos utilizar fracciones parciales. A continuación, vamos a obtener una serie de zeta-como sumas. Algunos de ellos vamos a reconocer de inmediato. Para el resto, podemos aplicar el teorema de los residuos.

$$\begin{align}\int_0^1 dx \frac{x^3 \log{x}}{(1+x^2)(1+x^3)} &= \frac12 \int_0^1 dx \left [\frac{1-x}{1+x^2} - \frac{1-x-x^2}{1+x^3} \right ] \log{x}\\ &= \frac12 \sum_{k=0}^{\infty} (-1)^k \int_0^1 dx \left (x^{2 k}-x^{2 k+1}-x^{3 k}+x^{3 k+1}+x^{3 k+2} \right ) \log{x}\\ &= -\frac12 \sum_{k=0}^{\infty} (-1)^k \left [ \frac1{(2 k+1)^2} - \frac1{(2 k+2)^2} - \frac1{(3 k+1)^2}\\ + \frac1{(3 k+2)^2}+\frac1{(3 k+3)^2}\right ] \\ &= \frac{5}{72} \frac{\pi^2}{12} - \frac12 G + \frac12 \sum_{k=0}^{\infty} (-1)^k \left [\frac1{(3 k+1)^2} - \frac1{(3 k+2)^2}\right ] \end{align} $$

donde $G$ es del catalán constante. Como para la suma:

$$\begin{align} \frac12 \sum_{k=0}^{\infty} (-1)^k \left [\frac1{(3 k+1)^2} - \frac1{(3 k+2)^2}\right ] &= \frac14 \sum_{k=-\infty}^{\infty} (-1)^k \left [\frac1{(3 k+1)^2} - \frac1{(3 k+2)^2}\right ] \\ &= -\frac{\pi}{4} \frac1{3^2} \left [\operatorname*{Res}_{z=-1/3} \frac{\csc{\pi z}}{(z+1/3)^2} \\- \operatorname*{Res}_{z=-2/3} \frac{\csc{\pi z}}{(z+2/3)^2} \right ]\\ &= \frac{\pi^2}{36} \left [\frac{\cos{\pi/3}}{\sin^2{\pi/3}} - \frac{\cos{2 \pi/3}}{\sin^2{2 \pi/3}} \right ]\\ &= \frac{\pi^2}{27}\end{align} $$

Por lo tanto

$$\int_0^1 dx \frac{x^3 \log{x}}{(1+x^2)(1+x^3)} = \frac{5 \pi^2}{864} - \frac12 G + \frac{\pi^2}{27} = \frac{37 \pi^2}{864} - \frac{G}{2}$$

y por último...

$$\int_0^{\infty} dx \frac{\log{(1+x)}}{(1+x^2)(1+x^3)} = \frac{G}{2} + \frac{\pi}{8} \log{2} - \frac{37 \pi^2}{864} \approx 0.307524\cdots$$

Answer 2

Podemos utilizar la integral paramétrica para evaluar la integral. Vamos $$ I(\alpha)=\int_0^\infty\frac{\ln(1+\alpha x)}{(1+x^2)(1+x^3)}dx. $$ Entonces es fácil ver \begin{eqnarray*} I'(\alpha)&=&\int_0^\infty\frac{x}{(1+\alpha x)(1+x^2)(1+x^3)}dx\\ &=&\pi\frac{(9+8\sqrt{3})\alpha^2+(8\sqrt{3}-9)}{(\alpha^2+1)(\alpha^2+\alpha+1)}-\frac{\alpha^3\ln\alpha}{(\alpha-1)(\alpha^2+1)(\alpha^2+\alpha+1)} \end{eqnarray*} y por lo tanto \begin{eqnarray*} I(1)&=&\int_0^1I'(\alpha)d\alpha\\ &=&\pi\int_0^1\frac{(9+8\sqrt{3})\alpha^2+(8\sqrt{3}-9)}{(\alpha^2+1)(\alpha^2+\alpha+1)}d\alpha+\int_0^1\frac{\alpha^3\ln\alpha}{(1-\alpha)(\alpha^2+1)(\alpha^2+\alpha+1)}d\alpha\\ &=&I_1+I_2. \end{eqnarray*} No es difícil de conseguir $$ I_1=\frac{\pi(5\pi+54\ln 2)}{432}. $$ Nota \begin{eqnarray*} \frac{\alpha^3}{(1-\alpha)(\alpha^2+1)(\alpha^2+\alpha+1)} &=&\frac{1}{6}\frac{1}{1-\alpha}-\frac{1}{2}\frac{1+\alpha}{1+\alpha^2}+\frac{1}{3}\frac{1+2\alpha}{1+\alpha^2+\alpha^2} \end{eqnarray*} y por lo tanto \begin{eqnarray*} I_2&=&\int_0^1\frac{\alpha^3\ln\alpha}{(1-\alpha)(\alpha^2+1)(\alpha^2+\alpha+1)}d\alpha\\ &=&\frac{1}{6}\int_0^1\frac{\ln\alpha}{1-\alpha}d\alpha-\frac{1}{2}\int_0^1\frac{(1+\alpha)\ln\alpha}{1+\alpha^2}d\alpha+\frac{1}{3}\int_0^1\frac{(1+2\alpha)\ln\alpha}{1+\alpha^2+\alpha^2}d\alpha \end{eqnarray*} Pero \begin{eqnarray*} \int_0^1\frac{\ln\alpha}{1-\alpha}d\alpha&=&\sum_{n=0}^\infty\int_0^1\alpha^n\ln\alpha d\alpha=-\sum_{n=0}^\infty\frac{1}{(n+1)^2}=-\frac{\pi^2}{6}\\ \int_0^1\frac{(1+\alpha)\ln\alpha}{1+\alpha^2}d\alpha&=&\int_0^1\frac{\ln\alpha}{1+\alpha^2}d\alpha+\int_0^1\frac{\alpha\ln\alpha}{1+\alpha^2}d\alpha\\ &=&\sum_{n=0}^\infty\int_0^1(-\alpha^2)^n\ln\alpha d\alpha+\sum_{n=0}^\infty\int_0^1\alpha(-\alpha^2)^n\ln\alpha d\alpha\\ &=&-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}-\sum_{n=0}^\infty\frac{(-1)^n}{4(n+1)^2}=-G-\frac{\pi^2}{48}\\ \int_0^1\frac{(1+2\alpha)\ln\alpha}{1+\alpha^2+\alpha^2}d\alpha&=&\int_0^1\frac{(1+\alpha-\alpha^2)\ln\alpha}{1-\alpha^3}d\alpha\\ &=&\int_0^1\frac{\ln\alpha}{1-\alpha^3}d\alpha+\int_0^1\frac{\alpha\ln\alpha}{1-\alpha^3}d\alpha-\int_0^1\frac{\alpha^2\ln\alpha}{1-\alpha^3}d\alpha\\ &=&\sum_{n=0}^\infty\int_0^\infty\alpha^{3n}\ln\alpha d\alpha+\sum_{n=0}^\infty\int_0^\infty\alpha^{3n+1}\ln\alpha d\alpha-\sum_{n=0}^\infty\int_0^\infty\alpha^{3n+2}\ln\alpha d\alpha\\ &=&-\sum_{n=0}^\infty\frac{1}{(3n+1)^2}-\sum_{n=0}^\infty\frac{1}{(3n+2)^2}+\frac{1}{9}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\\ &=&-\sum_{n=0}^\infty\frac{1}{(3n+1)^2}-\sum_{n=0}^\infty\frac{1}{(3n+2)^2}-\frac{1}{9}\sum_{n=0}^\infty\frac{1}{(n+1)^2}+\frac{2}{9}\sum_{n=0}^\infty\frac{1}{(n+1)^2}\\ &=&-\frac{\pi^2}{6}+\frac{2}{9}\frac{\pi^2}{6}=\frac{\pi^2}{54}. \end{eqnarray*} Poner todo junto, tenemos $$ I(1)=\frac{G}{2}-\frac{37\pi}{864}+\frac{1}{8}\pi\ln 2. $$

Answer 3

Sugerencia

Este sería probablemente el trabajo. Deje $u = \ln(1+x)$ $du = \frac{dx}{1+x}$ y el factor de la parte inferior $1+x^3 = (1+x)(\ldots)$.