Probar

Question

Integrar

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Por el teorema de Frullani

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¿Cómo podemos probar (4)?

Answer 1

$\newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert}$

\begin{align} &\color{#f00}{\int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{2x}}\, {\dd x \over x}} = \int_{0}^{\infty}{\expo{-x} - \expo{-2x} \over 1 + \expo{-2x}}\,\expo{-x}\, {\dd x \over x}\ \stackrel{\phantom{AA}t\ =\ \expo{-x}}{=}\ -\int_{0}^{1}{t - t^{2} \over 1 + t^{2}}\,{\dd t \over \ln\pars{t}} \\[3mm] = &\ \int_{0}^{1}{t - t^{2} \over 1 + t^{2}}\int_{0}^{\infty}t^{\mu}\,\dd\mu\,\dd t = \int_{0}^{\infty}\int_{0}^{1}{t^{\mu + 1} - t^{\mu + 2} \over 1 + t^{2}} \,\dd t\,\dd\mu = \half\int_{0}^{\infty}\int_{0}^{1}{t^{\mu/2} - t^{\mu/2 + 1/2} \over 1 + t} \,\dd t\,\dd\mu \\[3mm] = &\ \half\int_{0}^{\infty}\bracks{% 2\int_{0}^{1}{t^{\mu/2} - t^{\mu/2 + 1/2} \over 1 - t^{2}}\,\dd t - \int_{0}^{1}{t^{\mu/2} - t^{\mu/2 + 1/2} \over 1 - t}\,\dd t}\,\dd\mu \\[3mm] = &\ \half\int_{0}^{\infty}\bracks{% \int_{0}^{1}{t^{\mu/4 - 1/2} - t^{\mu/4 - 1/4} \over 1 - t}\,\dd t - \int_{0}^{1}{t^{\mu/2} - t^{\mu/2 + 1/2} \over 1 - t}\,\dd t}\,\dd\mu \\[3mm] = & \half\int_{0}^{\infty}\bracks{% \Psi\pars{{\mu \over 4} + {3 \over 4}} - \Psi\pars{{\mu \over 4} + \half} + \Psi\pars{{\mu \over 2} + 1} - \Psi\pars{{\mu \over 2} + {3 \over 2}}}\,\dd\mu \\[3mm] = & \left.\half \ln\pars{{\Gamma^{4}\pars{\mu/4 + 3/4}\Gamma^{2}\pars{\mu/2 + 1} \over \Gamma^{4}\pars{\mu/4 + 1/2}\Gamma^{2}\pars{\mu/2 + 3/2}}} \right\vert_{\ 0}^{\infty} = \bracks{-\,\half\ln\pars{2}} - \bracks{% -\,\half\,\ln\pars{2\pi} + \ln\pars{{\Gamma\pars{3/4} \over \Gamma\pars{5/4}}}} \\[3mm] = &\ \ln\pars{\root{\pi}\,{\Gamma\pars{5/4} \over \Gamma\pars{3/4}}} \end{align}

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Sin embargo, $$ \raíz{\pi}\,{\Gamma\pars{5/4} \over \Gamma\pars{3/4}} = \raíz{\pi}\,{1 \over 4}\,\Gamma\pars{1/4}\, {1 \over \pi/\bracks{\Gamma\pars{1/4}\sin\pars{\pi/4}}} ={\Gamma^{2}\pars{1/4} \más de 4\raíz{2\pi}} $$ y $$ \color{#f00}{\int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{2x}}\, {\dd x \sobre x}} = \color{#f00}{\ln\pars{{\Gamma^{2}\pars{1/4} \más de 4\raíz{2\pi}}}} \aprox 0.2708 $$

Answer 2

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ahora es suficiente considerar el límite como$$\prod_{n=1}^{2N}\left(\frac{2n+1}{2n}\right)^{(-1)^{n+1}} = \prod_{m=1}^{N}\frac{(4m)(4m-1)}{(4m+1)(4m-2)}=\frac{\sqrt{\pi }\, \Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{3}{4}+N\right) \Gamma(1+N)}{\Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{2}+N\right) \Gamma\left(\frac{5}{4}+N\right)} $ para recuperar la expresión deseada.