Pruebas $\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$

Question

Actualmente me encuentro con una integral que me resulta difícil evaluarla. Por favor, ayúdenme a evaluarla. Muchas gracias.

$$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$$

donde $\gamma$ es La constante de Euler-Mascheroni .

Answer 1

Una generalización: \begin{align} \int^\infty_0\frac{x}{(x^2+w^2)(1+e^{2\pi x})}{\rm d}x\tag1 =&\int^\infty_0\frac{xe^{-x}}{(x^2+4\pi^2w^2)(1+e^{-x})}{\rm d}x\\ \tag2 =&\int^\infty_0\frac{x}{x^2+4\pi^2w^2}\left(\sum^\infty_{n=1}(-1)^{n-1}e^{-nx}\right){\rm d}x\\ \tag3 =&\int^\infty_0xe^{-x}\left(\sum^\infty_{n=1}\frac{(-1)^{n-1}}{x^2+4n^2\pi^2w^2}\right){\rm d}x\\ =&\int^\infty_0\frac{e^{-x}}{2x}-\frac{e^{-x}}{4w}\mathrm{csch}\left(\frac{x}{2w}\right){\rm d}x\tag4\\ =&-\frac{1}{2}\int^1_0\frac{1}{\ln{x}}+\frac{x^{1/2w}}{w(1-x^{1/w})}\ {\rm d}x\tag5\\ =&-\frac{1}{2}\int^1_0\frac{x^{w-1}}{\ln{x}}+\frac{x^{w-1/2}}{1-x}{\rm d}x\tag6\\ =&-\frac{1}{2}\int^0_\infty\int^1_0x^{t+w-1}+\frac{x^{t+w-1/2}\ln{x}}{1-x}\ {\rm d}x\ {\rm d}t\tag7\\ =&-\frac{1}{2}\int^0_\infty\frac{1}{t+w}-\psi_1\left(t+w+\frac{1}{2}\right)\ {\rm d}t\tag8\\ =&\boxed{\large{\color{blue}{\displaystyle\frac{1}{2}\psi_0\left(w+\frac{1}{2}\right)-\frac{1}{2}\ln{w}}}}\\ \end{align}

Explicación:

$(1)$ : Sustituido $x\mapsto\dfrac{x}{2\pi}$ .
$(2)$ : Ampliado $\dfrac{1}{1+e^{-x}}$ como una serie geométrica.
$(3)$ : Sustituido $x\mapsto\dfrac{x}{n}$ .
$(4)$ : Se puede demostrar que $\displaystyle\sum^\infty_{n=-\infty}\frac{(-1)^n}{x^2+4n^2\pi^2w^2}=\frac{1}{2wx}\mathrm{csch}\left(\frac{x}{2w}\right)$ utilizando el teorema del residuo.
$(5)$ : Sustituido $x\mapsto-\ln{x}$ .
$(6)$ : Sustituido $x\mapsto x^w$ .
$(7)$ : Utilizó el hecho de que $\displaystyle\frac{1}{\ln{x}}=\int^0_\infty x^{t}\ {\rm d}t$ .
$(8)$ : Utiliza la representación integral de la función poligamma.

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El Integral:

Sea $w=1$ para obtener $$\int^\infty_0\frac{x}{(x^2+1)(e^{2\pi x}+1)}{\rm d}x=\frac{1}{2}\psi_0\left(\frac32\right)=\boxed{\large{\color{red}{\displaystyle 1-\frac{\gamma}{2}-\ln{2}}}}$$

Answer 2

He aquí una forma elemental de evaluar la integral sin utilizar funciones especiales ni fórmulas avanzadas. Observa que $$\int_0^\infty e^{-y}\sin(xy)\;\mathrm dy=\frac{x}{1+x^2}$$ Por lo tanto, tenemos \begin{align} \int_0^\infty \frac{x}{\left(1+x^2\right)\left(e^{2\pi x}+1\right)}\,\mathrm dx&=\int_0^\infty\int_0^\infty \frac{e^{-y}\sin(xy)}{e^{2\pi x}+1}\mathrm dy\;\mathrm dx\\[7pt] &=\int_0^\infty e^{-y}\int_0^\infty \frac{\sin(xy)}{e^{2\pi x}+1}\mathrm dx\;\mathrm dy \end{align} Se puede utilizar la siguiente técnica para evaluar la integral interna \begin{align} \int_0^\infty \frac{\sin(xy)}{e^{2\pi x}+1}\mathrm dx&=\int_0^\infty \frac{e^{-2\pi x}\sin(xy)}{1+e^{-2\pi x}}\mathrm dx\\[7pt] &=\sum_{k=1}^\infty(-1)^{k-1}\int_{0}^{\infty} e^{-2\pi kx}\sin (xy)\;\mathrm dx\\[7pt] &=\sum_{k=1}^\infty(-1)^{k-1}\frac{y}{4\pi^2k^2+y^2}\\[7pt] &=\frac{1}{4}\left[\frac{2}{y}-\operatorname{csch}\left(\frac{y}{2}\right)\right] \end{align} entonces obtenemos \begin{align} \int_0^\infty \frac{x}{\left(1+x^2\right)\left(e^{2\pi x}+1\right)}\,\mathrm dx&=\frac{1}{4}\int_{0}^{\infty} e^{- y}\left[\frac{2}{y}-\operatorname{csch}\left(\frac{y}{2}\right)\right]\mathrm dy\\[7pt] &=\frac{1}{2}\int_{0}^{\infty} e^{- y}\left[\frac{1}{y}-\frac{e^{-y/2}}{1-e^{-y}}\right]\mathrm dy\\[7pt] &=\frac{1}{2}\int_{0}^{\infty} e^{- y}\left[\frac{1}{y}-\frac{1}{1-e^{-y}}+\frac{e^{y/2}}{1+e^{y/2}}\right]\mathrm dy\\[7pt] &=\frac{1}{2}\underbrace{\int_{0}^{\infty} \left[\frac{e^{- y}}{y}-\frac{1}{e^{y}-1}\right]\mathrm dy}_{\large\color{blue}{(*)}}+\frac{1}{2}\int_{0}^{\infty} \frac{e^{-y/2}}{1+e^{y/2}}\mathrm dy\\[7pt] &=-\frac{\gamma}{2}+\frac{1}{2}\int_{0}^{\infty} \left[e^{-y/2}-\frac{e^{-y/2}}{1+e^{-y/2}}\right]\mathrm dy\\[7pt] &=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large1-\frac{\gamma}{2}-\ln2}} \end{align} donde $\color{blue}{(*)}$ es la representación integral de la constante de Euler-Mascheroni .

Answer 3

Pista: utiliza la segunda fórmula de Binet http://mathworld.wolfram.com/BinetsLogGammaFormulas.html . $$\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx$$ $$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} dx-2\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{4\pi x}-1)} dx$$ $$=\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}-1)} dx-\frac{1}{2}\int_{0}^{\infty}\frac{x}{((x/2)^2+1)(e^{2\pi x}-1)} dx$$

Ahora, vemos que

$$\frac{\partial}{\partial z}\left(\int_0^{\infty} \frac{\arctan(x/z)}{e^{2 \pi x}-1} \ dx\right)=-\int_0^{\infty}\frac{x}{\displaystyle \left(e^{2 \pi x}-1\right) z^2 \left(\frac{x^2}{z^2}+1\right)} \ dx$$

¿Puedes seguir desde aquí?

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} + 1}}\,\dd x} \\[5mm]&=\int_{0}^{\infty}{x \over \pars{x^{2} + 1}} \pars{{1 \over \expo{2\pi x} + 1} - {1 \over \expo{2\pi x} - 1}}\,\dd x +\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} - 1}} \\[5mm]&=-2\int_{0}^{\infty} {x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{4\pi x} - 1}} +\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} - 1}} \\[5mm]&=-2\int_{0}^{\infty} {x\,\dd x \over \pars{x^{2} + 4}\pars{\expo{2\pi x} - 1}} +\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} - 1}} \end{align}

Con identidad ${\bf 6.3.21}$ : $$ \int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + z^{2}}\pars{\expo{2\pi x} - 1}} =\half\bracks{\ln\pars{z} - {1 \over 2z} - \Psi\pars{z}} $$ conseguiremos

\begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + 1}\pars{\expo{2\pi x} + 1}}\,\dd x} \\[5mm]&=-2\braces{% \half\bracks{\ln\pars{z} - {1 \over 2z} - \Psi\pars{z}}_{z\ =\ 2}} +\half\bracks{\ln\pars{z} - {1 \over 2z} - \Psi\pars{z}}_{z\ =\ 1} \\[5mm]&=-\ln\pars{2} + {1 \over 4} +\ \overbrace{\Psi\pars{2}}^{\dsc{\Psi\pars{1} + 1}} -{1 \over 4} - \half\ \overbrace{\Psi\pars{1}}^{\dsc{-\gamma}} =\color{#66f}{\large 1 - {\gamma \over 2} - \ln\pars{2}} \end{align}

Answer 5

Doy una prueba de
$$ \int^{\infty}_{0}\frac{x}{x^2+a^2}\frac{1}{e^{2\pi x}-1}dx=-\frac{1}{2}\psi(a)-\frac{1}{4a}+\frac{\log(a)}{2} $$

Utilizamos las transformadas senoidales de Fourier: $$ \frac{x}{x^2+a^2}\leftrightarrow \sqrt{\frac{\pi}{2}}e^{-aw} $$ y $$ \frac{1}{e^{2\pi x}-1}\leftrightarrow \frac{1}{\sqrt{2 \pi}}\frac{-1+w\coth(w/2)}{2w} $$ Entonces podemos escribir $$ I=\int^{\infty}_{0}\frac{x}{x^2+a^2}\frac{1}{e^{2\pi x}-1}dx= $$ $$ =\int^{\infty}_{0}\frac{w\coth(w/2)-2}{4w}e^{-a w}dw= $$ $$ =-\frac{1}{4a}-\frac{1}{2}\frac{\Gamma'(a)}{\Gamma(a)}+\frac{\log(a)}{2}. $$ Desde $$ \int^{\infty}_{h}\frac{e^{-a w}}{w}dw=\int^{\infty}_{ah}\frac{e^{-t}}{t}dt=-e^{-ah}\log(ah)+\int^{\infty}_{ah}e^{-t}\log(t)dt\textrm{ : }\textrm{(1)} $$ y $$ \frac{1}{4}\int^{\infty}_{h}\coth(w/2)e^{-aw}dw=\frac{1}{4}\int^{e^{-h}}_0\frac{t^{a-1}(1+t)}{1-t}dt= $$ $$ =\frac{1}{4}\int^{e^{-h}}_{0}\left(\frac{t^{a-1}}{1-t}-\frac{1}{1-t}\right)dt+\frac{1}{4}\int^{e^{-h}}_{0}\left(\frac{t^a}{1-t}-\frac{1}{1-t}\right)dt-\frac{1}{2}\log\left(1-e^{-h}\right)= $$ $$ =\frac{1}{4}\sum^{\infty}_{k=0}\left(\frac{e^{-(k+a)h}}{k+a}-\frac{e^{-(k+1)h}}{k+1}\right) +\frac{1}{4}\sum^{\infty}_{k=0}\left(\frac{e^{-(k+a+1)h}}{k+a+1}-\frac{e^{-(k+1)h}}{k+1}\right)- $$ $$ -\frac{1}{2}\log\left(1-e^{-h}\right)\textrm{ : }\textrm{(2)} $$ A partir de (1) y (2) tenemos (tomando $h\rightarrow 0$ ): $$ I=\frac{-1}{4}\{\gamma+\psi(a)+\gamma+\psi(a+1)\}-\frac{1}{2}\lim_{h\rightarrow 0}\left(\log(1-e^{-h})-e^{-ah}\log(ah)\right)+\frac{\gamma}{2}= $$ $$ =-\frac{1}{2}\psi(a)-\frac{1}{4a}+\frac{\log(a)}{2}. $$ Dónde hemos utilizado $\int^{\infty}_{0}e^{-t}\log(t)dt=-\gamma$ .