Me gustaría evaluar esta integral. PS
$$\int_{0}^{1}\frac{x(1-x)^2}{1+x+x^2}\frac{\mathrm dx}{\ln x}$
PS
PS
Hice una sustitución inútil, esperaba obtener una integral más simple.
Respuestas
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Considere la función
$$ I(s) = \int_{0}^{1} \frac{x^{s+1} (1 - x)^2}{1+x+x^2}\,\frac{\mathrm{d}x}{\log x}. $$
Entonces
\begin{align*} I'(s) &= \int_{0}^{1} \frac{x^{s+1} (1 - x)^3}{1-x^3}\,\mathrm{d}x \\ &= \frac{1}{3}\int_{0}^{1} \frac{u^{(s-1)/3} (1 - u^{1/3})^3}{1-u}\,\mathrm{d}u \tag{%#%#%} \\ &= \frac{1}{3} \sum_{k=0}^{3} \binom{3}{k} (-1)^{k-1} \int_{0}^{1} \frac{1-u^{(s+k-1)/3}}{1-u}\,\mathrm{d}u \\ &= \frac{1}{3} \sum_{k=0}^{3} \binom{3}{k} (-1)^{k-1} \psi((s+k+2)/3), \end{align*}
donde hemos utilizado la identidad de $x=u^{1/3}$. También, $\sum_{k=0}^{3}\binom{3}{k}(-1)^k = 0$ es la función digamma y hemos utilizado la identidad que $\psi$ en el último paso. Junto con $\int_{0}^{1} \frac{1-u^z}{1-u} \, \mathrm{d}z = \gamma + \psi(z+1)$, obtenemos
\begin{align*} I(0) &= -\lim_{R\to\infty}\int_{0}^{R} I'(s) \, \mathrm{d}s \\ &= \lim_{R\to\infty}\sum_{k=0}^{3} \binom{3}{k} (-1)^{k} \left[ \log\Gamma((R+k+2)/3) - \log\Gamma((k+2)/3) \right] \\ &= - \sum_{k=0}^{3} \binom{3}{k} (-1)^{k} \log\Gamma((k+2)/3) \\ &= \log(18) - 3\log\Gamma(1/3). \end{align*}
Felix Marin
Puntos
32763
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{x\pars{1 - x}^{2} \over 1 + x + x^{2}}\,{\dd x \over \ln\pars{x}}} = \int_{0}^{1}{x - 2x^{2} + x^{3} \over 1 - x^{3}}\ \overbrace{\int_{0}^{1}x^{t}\,\dd t}^{\ds{-\,{1 - x \over \ln\pars{x}}}}\ \,\dd x \\[5mm] = &\ -\int_{0}^{1}\int_{0}^{1}{x^{1 + t} - 2x^{2 + t} + x^{3 + t} \over 1 - x^{3}}\,\dd x\,\dd t \\[5mm] \stackrel{x^{3}\ \mapsto\ x}{=}\,\,\,& -\,{1 \over 3}\int_{0}^{1}\int_{0}^{1}{x^{-1/3 + t/3} - 2x^{t/3} + x^{1/3 + t/3} \over 1 - x}\,\dd x\,\dd t \\[5mm] = &\ {1 \over 3}\int_{0}^{1}\bracks{% \int_{0}^{1}{1 - x^{t/3 - 1/3} \over 1 - x}\,\dd x - 2\int_{0}^{1}{1 - x^{t/3} \over 1 - x}\,\dd x + \int_{0}^{1}{1 - x^{t/3 + 1/3} \over 1 - x}\,\dd x}\dd t \end{align}
Con la Identidad de $\mathbf{\color{black}{6.3.22}}$ en a & S Tabla :
\begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{x\pars{1 - x}^{2} \over 1 + x + x^{2}}\,{\dd x \over \ln\pars{x}}} \\[5mm] = &\ {1 \over 3}\int_{0}^{1}\bracks{% \Psi\pars{{t \over 3} + {2 \over 3}} - 2\Psi\pars{{t \over 3} + 1} + \Psi\pars{{t \over 3} + {4 \over 3}}}\dd t \end{align}
$\ds{\Psi}$ es la Función Digamma.
A continuación, \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{x\pars{1 - x}^{2} \over 1 + x + x^{2}}\,{\dd x \over \ln\pars{x}}} \\[5mm] = &\ \left. \ln\pars{\Gamma\pars{{t \over 3} + {2 \over 3}}} - 2\ln\pars{\Gamma\pars{{t \over 3} + 1}} + \ln\pars{\Gamma\pars{{t \over 3} + {4 \over 3}}} \,\right\vert_{\ 0}^{\ 1} \\[8mm] = &\ \ln\pars{\Gamma\pars{1}} - 2\ln\pars{\Gamma\pars{4 \over 3}} + \ln\pars{\Gamma\pars{5 \over 3}} -\ln\pars{\Gamma\pars{2 \over 3}} \\[2mm] &\ + 2\ln\pars{\Gamma\pars{1}} - \ln\pars{\Gamma\pars{4 \over 3}} \\[8mm] = &\ -3\ln\pars{\Gamma\pars{4 \over 3}} + \ln\pars{\Gamma\pars{5 \over 3}} - \ln\pars{\Gamma\pars{2 \over 3}} \label{1}\tag{1} \end{align} donde $\ds{\Gamma}$ es el La Función Gamma.
Con $\ds{\Gamma}$ De La Recurrencia De La Fórmula \begin{equation} \left\{\begin{array}{rcl} \ds{-3\ln\pars{\Gamma\pars{4 \over 3}}} & \ds{=} & \ds{\overbrace{-3\ln\pars{1 \over 3}}^{\ds{\ln\pars{27}}}\ -\ 3\ln\pars{\Gamma\pars{1 \over 3}}} \\[2mm] \ds{\ln\pars{\Gamma\pars{5 \over 3}}} & \ds{=} & \ds{\phantom{-3\,}\ln\pars{2 \over 3} + \ln\pars{\Gamma\pars{2 \over 3}}} \end{array}\right. \label{2}\etiqueta{2} \end{equation}
\eqref{1} y \eqref{2} $\ds{\implies}$ $$ \bbox[10px,#ffd]{\int_{0}^{1}{x\pars{1 - x}^{2} \over 1 + x + x^{2}}\,{\dd x \sobre \ln\pars{x}}} = \bbx{\ln\pars{18 \sobre \Gamma^{3}\pars{1/3}}} \approx -0.0659 $$
