Integral de la $I=\int_0^\infty \frac{e^{\alpha x}-e^{\beta x}}{x(e^{\alpha x}+1)(e^{\beta x}+1)}dx, \ \ \alpha>\beta>0. $

Question

$$ I(\alpha,\beta)=\int_0^\infty \frac{e^{\alpha x}-e^{\beta x}}{x(e^{\alpha x}+1)(e^{\beta x}+1)}dx, \ \ \alpha>\beta>0. $$ Estoy tratando de resolver esta integral. Esto es de la vieja escuela de alto días en Bulgaria, aunque no puedo encontrar las soluciones más. Gracias

Answer 1

Dividir el integrando:

$$\frac{e^{\alpha x}-e^{\beta x}} {e^{\alpha x}+1)(e^{\beta x}+1)}= \frac{1} {e^{\beta x}+1)}-\frac{1} {e^{\alpha x}+1)}$$

Reconocer esto como una integral de una función, evaluados en los límites de $\alpha$$\beta$. Qué función? Así, tomar la derivada de más de $\beta$ saber:

$$\frac{d}{dy}\left(\frac{1}{x}\frac{1}{(e^{y x}+1)}\right)=-\frac{e^{xy}}{(e^{y x}+1)^2}$$

Su integral se convierte en

$$I=\int_0^\infty\int_{\beta}^\alpha \frac{e^{xy} dy\,dx}{(e^{xy}+1)^2}$$

Intercambiar el orden de integración:

$$I=\int_{\beta}^\alpha\int_0^\infty \frac{e^{xy} dx\,dy}{(e^{xy}+1)^2}$$

El interior de la integral $$\int_0^\infty \frac{e^{xy} dx}{(e^{xy}+1)^2}$$ Se convierte en $$\frac{1}{y}\int_0^\infty \frac{e^{u} du}{(e^{u}+1)^2}$$ El uso de $v=e^u+1$: $$\frac{1}{y}\int_2^\infty \frac{dv}{v^2}=\frac12\frac{1}{y}$$

Ahora el exterior de la integral es trivial:

$$I=\frac12\int_{\beta}^\alpha \frac{dy}{y}=\frac12\ln\frac{\alpha}{\beta}$$

Answer 2

$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm I}\pars{\alpha,\beta} =\int_{0}^{\infty}{\expo{\alpha x} -\expo{\beta x} \over x\pars{\expo{\alpha x} + 1}\pars{\expo{\beta x} + 1}}\,\dd x:\ {\large ?}\,, \qquad \alpha > \beta > 0}$.

\begin{align} \color{#00f}{\large{\rm I}\pars{\alpha,\beta}}&=\int_{0}^{\infty} \pars{{1 \over \expo{\beta x} + 1} - {1 \over \expo{\alpha x} + 1}}\,{\dd x \over x} \\[3mm]&= \int_{0}^{\infty}\braces{\half\bracks{1 - \tanh\pars{\beta x \over 2}} - \half\bracks{1 - \tanh\pars{\alpha x \over 2}}}\,{\dd x \over x} \\[3mm]&=\lim_{\Lambda \to \infty}\braces{% \int_{0}^{\Lambda}\half\bracks{1 - \tanh\pars{\beta x \over 2}}\,{\dd x \over x} - \int_{0}^{\Lambda}\half\bracks{1 - \tanh\pars{\alpha x \over 2}}\,{\dd x \over x}} \\[3mm]&=\half\lim_{\Lambda \to\infty}\bracks{% \int_{0}^{\Lambda}\tanh\pars{\alpha x \over 2}\,{\dd x \over x} -\int_{0}^{\Lambda}\tanh\pars{\beta x \over 2}\,{\dd x \over x}} \\[3mm]&=\half\lim_{\Lambda \to\infty}\bracks{% \int_{0}^{\alpha\Lambda/2}{\tanh\pars{x} \over x}\,\dd x -\int_{0}^{\beta\Lambda/2}{\tanh\pars{x} \over x}\,\dd x} \\[3mm]&=\half\lim_{\Lambda \to\infty}\left\lbrace% \bracks{\ln\pars{\alpha\Lambda \over 2}\tanh\pars{\alpha\Lambda \over 2} -\int_{0}^{\alpha\Lambda/2}\ln\pars{x}\sech^{2}\pars{x}\,\dd x}\right. \\[3mm]&\left.\phantom{\half\lim_{\Lambda \to\infty}\left\lbrace\right.} \mbox{}-\bracks{\ln\pars{\beta\Lambda \over 2}\tanh\pars{\beta\Lambda \over 2} -\int_{0}^{\beta\Lambda/2}\ln\pars{x}\sech^{2}\pars{x}\,\dd x}\right\rbrace \\[3mm]&=\half\lim_{\Lambda \to \infty}\bracks{% \ln\pars{\alpha\Lambda \over 2} - \ln\pars{\beta\Lambda \over 2}} =\color{#00f}{\large\half\ln\pars{\alpha \over \beta}} \end{align}