Una posible forma cerrada?

Question

Tiene una forma cerrada?

$$\int _{0}^{\infty }\!{\frac {x\cos \left( x \right) -\sin \left( x \right) }{{x} \left( {{\rm e}^{x}}-1 \right) }}{dx}$$

EDIT: no hay necesidad de contestar, acabo de encontrar la forma cerrada. Gracias!

Answer 1

Es esto lo que han encontrado?

$$\int _{0}^{\infty }\!{\frac {x\cos \left( x \right) -\sin \left( x \right) }{{x} \left( {{\rm e}^{x}}-1 \right) }}{dx} = \frac{\pi}{2}+\arg\left(\Gamma\left(i\right)\right)-\Re\left(\psi_0\left(i\right)\right).$$

Aquí $\arg$ es el argumento complejo, $\Re$ es la parte real de un número complejo, $\Gamma$ es la función gamma, $\psi_0$ es la función digamma, $i$ es la unidad imaginaria y $\pi$ es también un famoso constante.

Answer 2

$\newcommand{\+}{^{\daga}}% \newcommand{\ángulos}[1]{\left\langle #1 \right\rangle}% \newcommand{\llaves}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \cima {= \cima \vphantom{\enorme}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\mitad}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\a la derecha\vert\,}% \newcommand{\cy}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert}$ \begin{align} &\int _{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x\pars{\expo{x} - 1}}\,\dd x = \Re\int _{0}^{\infty} \bracks{\expo{\ic x} - {\sin\pars{x} \over x}}\,{1 \over \expo{x} - 1}\,\dd x \\[3mm]&= \Re\int _{0}^{\infty} \bracks{\expo{\pars{\ic - 1}x} - {\sin\pars{x}\expo{-x} \over x}}\, {\dd x \over 1 - \expo{x}} = \Re\int _{0}^{\infty} \bracks{\expo{-\pars{1 - \ic}x} - {\sin\pars{x}\expo{-x} \over x}}\, \sum_{\ell = 0}^{\infty}\expo{-x}\,\dd x \\[3mm]&= \Re\sum_{\ell = 0}^{\infty}\int _{0}^{\infty} \bracks{\expo{-\pars{\ell + 1 - \ic}x} - {\sin\pars{x}\expo{-\pars{\ell + 1}x} \over x}}\,\dd x \end{align}

También $$ \totald{}{\mu}\int_{0}^{\infty}{\sin\pars{x}\expo{-\mu x} \over x}\,\dd x = -\Im\int_{0}^{\infty}\expo{\pars{\ic - \mu}x}\,\dd x =-\Im\pars{1 \over \mu \ic} = -\,{1 \over \mu^{2} + 1} $$ $$ \int_{0}^{\infty}{\sin\pars{x}\expo{-\pars{\ell + 1}x} \over x}\,\dd x = {\pi \over 2} - \int_{0}^{\ell + 1}{\dd\mu \\mu^{2} + 1} ={\pi \over 2} - \arctan\pars{\ell + 1} = \arctan\pars{1 \over \ell + 1} $$

\begin{align} &\color{#00f}{\large\int _{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x\pars{\expo{x} - 1}}\,\dd x = \Re\sum_{\ell = 0}^{\infty} \bracks{{1 \over \ell + 1 - \ic} - \arctan\pars{1 \over \ell + 1}}} \end{align}