Interesante integral: $\int_0^1{\frac{nx^{n-1}}{x+1}}dx$

Question

Encontrar el valor de %#% $ #%

No tuvo mucha suerte al mismo tiempo de integración. También intenté distinguir w.r.t n pero todavía no podía llegar a cualquier punto. Necesitan ayuda.

Answer 1

Puesto $y=1+x$ y la integral se convierte en $$ \int_{1}^2 \frac{n(y-1) ^ {n-1}} {y} \, dy = \int_1^2 \sum_{k=0}^n n\binom {n-1} {k} (-1) ^ {n-k-1} y ^ {k-1} \, dy = \left [n (-1) ^ {n-1} \log {y} + \sum_{k=1}^n \frac{n}{k} \binom{n-1}{k} (-1) ^ {n-k-1} y ^ k \right]_1^2 \\ = n-(1) ^ \log {n-1} {2} + \sum_{k=1}^n \frac{n}{k} \binom{n-1}{k} (-1)^{n-k-1}(2^k-1). $$

Answer 2

SUGERENCIA:

Si $\displaystyle I_n=\int_0^1\dfrac{nx^{n-1}}{x+1}dx,$

$$I_{m+1}+I_m=\int_0^1\dfrac{(m+1)x^m+mx^{m-1}}{x+1}dx$$ $$=m\int_0^1x^{m-1}\ dx+\int_0^1\dfrac{x^m}{1+x}dx=1+\int_0^1\dfrac{x^m}{1+x}dx$$

Otra vez si $\displaystyle J_m=\int_0^1\dfrac{x^m}{1+x}dx,$

$$J_{m+1}+J_m=?$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{n\,x^{n - 1} \over x + 1}\,\dd x & = n\int_{0}^{1}{x^{n - 1} - x^{n} \over 1 - x^{2}}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\,n\int_{0}^{1}{x^{n/2 - 1} - x^{n/2 - 1/2} \over 1 - x}\,\dd x \\[5mm] & = {1 \over 2}\,n\bracks{% \int_{0}^{1}{1 - x^{n/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}{1 - x^{n/2 - 1} \over 1 - x}\,\dd x} \\[5mm] & = \bbx{{1 \over 2}\,n\pars{H_{n/2 - 1/2} - H_{n/2 - 1}}} \end{align}

donde $\ds{H_{z}}$ es el Número Armónico.

Answer 4

Podemos escribir la integral como $$ F(n) = \int_0^{\;1} {\frac{{nx^{n - 1} }}{{1 + x}}dx} = \int_0^{\;1} {\frac{1}{{1 + x}}dx^n } = \int_0^{\;1} {\left( {1 - \frac{x}{{1 + x}}} \right)dx^n } = 1 - \int_0^{\;1} {\frac{x}{{1 + x}}dx^n } $$

$F(n+1)$ $$ F(n + 1) = \int_0^{\;1} {\frac{{\left( {n + 1} \right)x^n }}{{1 + x}}dx} = \int_0^{\;1} {\frac{x}{{1 + x}}dx^n } + \int_0^1 {\frac{{x^n }}{{1 + x}}dx} = 1 - F(n) + \frac{1}{{n + 1}}F(n + 1) $$ es decir: $$ \begin{array}{l} \left( {1 - \frac{1}{{n + 1}}} \right)F(n + 1) = 1 - F(n) \\ \frac{{F(n + 1)}}{{n + 1}} = \frac{1}{n} - \frac{{F(n)}}{n} \\ \left( { - 1} \right)^{\,n} \frac{{F(n + 1)}}{{n + 1}} = \frac{{\left( { - 1} \right)^{\,n} }}{n} + \left( { - 1} \right)^{\,n - 1} \frac{{F(n)}}{n} \\ \end{array} $$

Así tenemos $$ \left\{ \begin{array}{l} G(n) = \left( { - 1} \right)^{\,n} \frac{{F(n + 1)}}{{n + 1}} \\ G(0) = F(1) = \ln 2 \\ G(n + 1) - G(n) = \frac{{\left( { - 1} \right)^{\,n + 1} }}{{n + 1}} \\ \end{array} \right. $$

lo que conduce a $$ \bbox[lightyellow] { \begin{array}{l} G(n) = \ln (2) + \sum\limits_{k = 0}^{n - 1} {\frac{{\left( { - 1} \right)^{\,k + 1} }}{{k + 1}}} = \left( { - 1} \right)^{\,n} \Phi \left( { - 1,1,n + 1} \right) \\ F(n) = n\;\Phi \left( { - 1,1,n} \right) \\ \end{array} }$$ donde $\Phi$ denota la Lerch Trascendente.

De hecho, por la representación integral de la $\Phi$ $$ \Phi \left( {z,s,a} \right)\mathop \equiv \límites^{def} \frac{1} {{\Gamma (s)}}\int_0^\infty {\frac{{t^{\s - 1} e^ {\, -\, t} }} {{1 - z\,e^{\, - \,t} }}dt} $$ y $$ \Phi \left( { - 1,1,n} \right) = \int_{t\, = \,0}^\infty {\frac{{e^{\, - n\,t} }} {{1 + \,e^{\, - \,t} }}dt} \quad \xrightarrow{{e^{\, - \,t} = x}}\quad \int_{x = 1}^0 {\frac{{x^{\,n - 1} }} {{1 + \,x}}dt} $$