Integral de la $\int_{1}^{\infty} \frac{\log^3 x}{x(x-1)} dx$

Question

¿Cómo puedo llegar a la forma cerrada de la expresión de la integral de la $$\displaystyle\int_{1}^{\infty} \dfrac{\log^3 x}{x(x-1)}dx$$

Probablemente la forma cerrada es $\dfrac{\pi^4}{15}$

Answer 1

El cambio de las variables de $x\leftarrow 1/x~$ muestra que $$\eqalign{ I&=\int_1^\infty\frac{\log^3x}{x(x-1)}dx\cr &=\int_0^1\frac{-\log^3x}{1-x}dx =-\sum_{n=0}^\infty \int_{0}^1^n \log^3 x\,dx\cr &=\sum_{n=0}^\infty \frac{6}{(n+1)^4}=6\zeta(4)=\frac{\pi^4}{15} } $$ cual es la respuesta deseada.

Edit. De hecho, en general, el cambio de las variables de $x=e^{-t}$ muestra que $$\eqalign{ \int_0^1^n\log^p(1/x)\,dx y=\int_0^\infty e^{-(n+1)t}t^pdt\cr &=\frac{1}{(n+1)^{p+1}}\int_0^\infty e^{-u}u^pdu\cr &= \frac{p!}{(n+1)^{p+1}}} $$

Answer 2

Utilizando un CAS, $$I_1=\displaystyle\int\dfrac{\log^3 x}{x(x-1)}dx=$$ $$6 \texto{Li}dimm_4(x)+3 \text{Li}_2(x) \log ^2(x)-6 \text{Li}_3(x) \log (x)-\frac{1}{4} \log ^4(x)+\log (1-x) \log ^3(x)$$ $$I_2=\displaystyle\int_{1}^{a} \dfrac{\log^3 x}{x(x-1)}dx=$$ $$6 \texto{Li}dimm_4(a)+3 \text{Li}_2(a) \log ^2(a)-6 \text{Li}_3(a) \log (a)-\frac{1}{4} \log ^4(a)+\log (1-a) \log ^3(a)-\frac{\pi ^4}{15}$$and, going to limit $$I_3=\displaystyle\int_{1}^{\infty} \dfrac{\log^3 x}{x(x-1)}dx=\frac{\pi ^4}{15}$$

Para valores grandes de $a$ $$I_2\simeq \frac{\pi ^4}{15}-\frac{\log ^3(a)+3 \log ^2(a)+6 \log (a)+6}{a}$$ For $un=10^3$, the value of the integral is $5.97352$ while the approximation leads to $5.97372$. For $un=10^4$, the value of the integral is $6.38423$ while the approximation leads to $6.38423$.

Answer 3

En primer lugar, hacer las sustituciones $x=\frac{1}{u}$, seguido por $u=1-z$, y luego integrar por partes dos veces. Sustituto $z=1-w$, luego de integrar por partes de nuevo:

$$\begin{align} \int_{1}^{\infty}\frac{\log^3{x}}{x(1-x)}\mathrm{d}x &=\int_{1}^{0}\frac{\log^3{\frac{1}{u}}}{\frac{1}{u}(1-\frac{1}{u})}\cdot\frac{(-\mathrm{d}u)}{u^2}\\ &=-\int_{0}^{1}\frac{\log^3{u}}{1-u}\mathrm{d}u\\ &=-\int_{0}^{1}\frac{\log^3{\left(1-z\right)}}{z}\mathrm{d}z\\ &=-\left[\ln{\left(z\right)}\log^3{\left(1-z\right)}\right]_{0}^{1}+\int_{0}^{1}\frac{3\ln{\left(z\right)}\ln^2{\left(1-z\right)}}{z-1}\mathrm{d}z\\ &=-3\int_{0}^{1}\frac{\ln{\left(z\right)}\ln^2{\left(1-z\right)}}{1-z}\mathrm{d}z\\ &=-3\left[\ln^2{\left(1-z\right)}\operatorname{Li}_{2}{\left(1-z\right)}\right]_{0}^{1}+3\int_{0}^{1}\frac{2\log{\left(1-z\right)}\operatorname{Li}_{2}{\left(1-z\right)}}{z-1}\mathrm{d}z\\ &=-6\int_{0}^{1}\frac{\log{\left(1-z\right)}\operatorname{Li}_{2}{\left(1-z\right)}}{1-z}\mathrm{d}z\\ &=-6\int_{0}^{1}\frac{\log{\left(w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w}\mathrm{d}w\\ &=-6\left[\log{\left(w\right)}\operatorname{Li}_{3}{\left(w\right)}\right]_{0}^{1}+6\int_{0}^{1}\frac{\operatorname{Li}_{3}{\left(w\right)}}{w}\mathrm{d}w\\ &=6\int_{0}^{1}\frac{\operatorname{Li}_{3}{\left(w\right)}}{w}\mathrm{d}w\\ &=6\operatorname{Li}_{4}{\left(1\right)}\\ &=\frac{\pi^4}{15}. \end{align}$$

Answer 4

$\newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert}$ $\ds{\int_{1}^{\infty}{\ln^{3}\pars{x} \over x\pars{x - 1}}\,\dd x:\ {\large ?}}$.

\begin{align}&\color{#66f}{\Large\int_{1}^{\infty}% {\ln^{3}\pars{x} \over x\pars{x - 1}}\,\dd x} =\int_{1}^{0}{\ln^{3}\pars{1/x} \over \pars{1/x - 1}/x}\, \pars{-\,{\dd x \over x^{2}}} =-\int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x \\[3mm]&=-\int_{0}^{1}\ln\pars{1 - x}\bracks{3\ln^{2}\pars{x}\,{1 \over x}}\,\dd x =3\int_{0}^{1}{\rm Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x \\[3mm]&=-3\int_{0}^{1}{\rm Li}_{2}\pars{x}\bracks{2\ln\pars{x}\,{1 \over x}} \,\dd x =-6\int_{0}^{1}{\rm Li}_{3}'\pars{x}\ln\pars{x}\,\dd x =6\int_{0}^{1}{\rm Li}_{3}\pars{x}\,{1 \over x}\,\dd x \\[3mm]&=6\int_{0}^{1}{\rm Li}_{4}'\pars{x}\,\dd x =6\,{\rm Li}_{4}\pars{1}=6\ \underbrace{\zeta\pars{4}}_{\ds{=\ \color{#c00000}{\pi^{4} \over 90}}}\ =\ 6\,{\pi^{4} \over 90}=\color{#66f}{\Large{\pi^{4} \over 15}} \approx {\tt 6.4939} \end{align}