Cómo mostrar que $2\int_{0}^{\infty}{\cosh(x)-1\over x(e^{ax}+1)} dx=\ln{4a\sin^2\left({\pi\over 2a}\right)\over \pi\sin\left({\pi\over a}\right)}?$

Question

¿Cómo podemos demostrar que

$$2\int_{0}^{\infty}{\cosh(x)-1\over x(e^{ax}+1)}\mathrm dx=\ln{4a\sin^2\left({\pi\over 2a}\right)\over \pi\sin\left({\pi\over a}\right)}?\tag1$$

Answer 1

Sugerencia. Se puede observar que la $$ \int_0^{1} \sinh (y\cdot x) \, dy=\frac{\cosh (x)-1}{x},\qquad x>0, $$ then one may obtain, for $>1$, $$ \begin{align} \int_{0}^{\infty}{\cosh(x)-1\over x(e^{ax}+1)}\mathrm dx&=\int_0^{1}dy\int_{0}^{\infty}{\sinh (y\cdot x)\over e^{ax}+1}\mathrm dx \\\\&=\frac{1}{2} \int_0^{1}\left(\frac{\frac{\pi}{a}}{\sin \left(\frac{\pi y}{a}\right)}-\frac{1}{y}\right)dy \\\\&=\frac{1}{2}\ln \left(\frac{2a}{\pi}\cdot\tan\left(\frac{\pi}{2a}\right)\right) \end{align} $$ donde hemos utilizado el hecho de que $$ \left(\frac{\ln \left(\tan \left(\frac{r}{2}\right)\right)}{b}\right)'_y=\frac{1+\tan^2 \left(\frac{r}{2}\right)}{2\tan \left(\frac{r}{2}\right)}=\frac{1}{2\sin\left(\frac{r}{2}\right)\cos\left(\frac{r}{2}\right)}=\frac1{\sin\left(b y\right)}. $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{2\int_{0}^{\infty}{\cosh\pars{x} - 1 \sobre x\pars{\expo{ax} + 1}}\,\dd x = \ln\pars{4a\sin^{2}\pars{\pi/\bracks{2a}} \over \pi\sin\pars{\pi/a}}:\ {\large ?}}$.

\begin{align} &2\int_{0}^{\infty}{\cosh\pars{x} - 1 \over x\pars{\expo{ax} + 1}}\,\dd x = \,\mrm{f}\pars{1 - {1 \over a}} + \mrm{f}\pars{1 + {1 \over a}} \label{1}\tag{1} \\[5mm] &\mbox{where}\quad \mrm{f}\pars{z} \equiv \int_{0}^{\infty}{\expo{-zax} - \expo{-ax} \over x\pars{1 + \expo{-ax}}}\,\dd x \label{2}\tag{2} \end{align}

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\begin{align} \mrm{f}\pars{z} & \equiv \int_{0}^{\infty}{\expo{-zax} - \expo{-ax} \over x\pars{1 + \expo{-ax}}}\,\dd x \,\,\,\,\,\,\stackrel{\pars{\substack{{\large\expo{-ax}\ =\ t}\\[0.5mm] {\large x\ =\ -\ln\pars{t}/a}}}\\[2mm] \mbox{}}{\Large=}\,\,\,\,\,\, -\int_{0}^{1}{t^{z - 1} - 1 \over \ln\pars{t}\pars{1 + t}}\,\dd t \\[5mm] & = \int_{0}^{1}{t^{z - 1} - 1 \over 1 - t^{2}}\,{t - 1 \over \ln\pars{t}}\,\dd t = \int_{0}^{1}{t^{z - 1} - 1 \over 1 - t^{2}}\int_{0}^{1}t^{s}\,\dd s\,\dd t = \int_{0}^{1}\int_{0}^{1}{t^{s + z - 1} - t^{s} \over 1 - t^{2}}\,\dd t\,\dd s \\[5mm] & \stackrel{t^{2}\ \mapsto\ t}{=}\,\,\, {1 \over 2}\int_{0}^{1}\int_{0}^{1} {t^{s/2 + z/2 - 1} - t^{s/2 - 1/2} \over 1 - t}\,\dd t\,\dd s \\[5mm] & = {1 \over 2}\int_{0}^{1}\pars{% \int_{0}^{1}{1 - t^{s/2 - 1/2} \over 1 - t}\,\dd t - \int_{0}^{1}{1 - t^{s/2 + z/2- 1} \over 1 - t}\,\dd t}\,\dd s \\[5mm] & = {1 \over 2}\int_{0}^{1}\bracks{% \Psi\pars{s + 1 \over 2} - \Psi\pars{s + z \over 2}} \dd s\qquad\pars{~\Psi:\ Digamma\ Function~} \\[5mm] & = \left.\ln\pars{\Gamma\pars{\pars{s + 1}/2} \over \Gamma\pars{\bracks{s + z}/2}} \right\vert_{\ s\ =\ 0}^{\ s\ =\ 1} = \ln\pars{{\Gamma\pars{1} \over \Gamma\pars{\bracks{1 + z}/2}}\, {\Gamma\pars{z/2} \over \Gamma\pars{1/2}}} \\[5mm] & \implies \bbx{\mrm{f}\pars{z} = \ln\pars{{1 \over \root{\pi}}\,{\Gamma\pars{z/2} \over \Gamma\pars{\bracks{1 + z}/2}}}} \end{align}

\eqref{1} se reduce a

\begin{align} &2\int_{0}^{\infty}{\cosh\pars{x} - 1 \over x\pars{\expo{ax} + 1}}\,\dd x = \ln\pars{{1 \over \root{\pi}} {\Gamma\pars{1/2 - 1/\bracks{2a}} \over \Gamma\pars{1 - 1/\bracks{2a}}}\, {1 \over \root{\pi}} {\Gamma\pars{1/2 + 1/\bracks{2a}} \over \Gamma\pars{1 + 1/\bracks{2a}}}} \\[5mm] = &\ \ln\pars{{1 \over \pi}\,{\pi \over \sin\pars{\pi\braces{1/2 + 1/\bracks{2a}}}}\, {1 \over \Gamma\pars{1 - 1/\bracks{2a}}\Gamma\pars{1/\bracks{2a}}/\bracks{2a}}} \\[5mm] = &\ \ln\pars{{2a \over \cos\pars{\pi/\bracks{2a}}}\, {1 \over \pi/\sin\pars{\pi/\bracks{2a}}}} = \ln\pars{{4a \over \pi}\,{\sin^{2}\pars{\pi/\bracks{2a}} \over 2\sin\pars{\pi/\bracks{2a}}\cos\pars{\pi/\bracks{2a}}}} \\[5mm] & = \bbx{\ln\pars{4a\sin^{2}\pars{\pi/\bracks{2a}} \over \pi\sin\pars{\pi/a}}} \end{align}

De hecho, una expresión más sencilla es $\bbx{\ds{\ln\pars{{2a \over \pi}\,\tan\pars{\pi \over 2a}}}}$.

Answer 3

Un enfoque alternativo. Por Frullani del teorema de $$ \int_{0}^{+\infty}\frac{\cosh(x)-1}{x}e^{-\mu x}\,dx = \log\mu-\frac{1}{2}\log(\mu^2-1)\tag{1}$$ para cualquier $\mu>1$. Mediante la expansión de $\frac{1}{e^{ax}+1}$ como una serie geométrica $$ \frac{1}{1+e^{ax}} = e^{-ax}-e^{-2a x}+e^{-3ax}-e^{-4ax}+\ldots \tag{2} $$ el original de la integral es igual a $$ \frac{1}{2}\sum_{k\geq 1}(-1)^{k+1}\left[\log(k^2a^2)-\log(k^2 a^2-1)\right]=\frac{1}{2}\,\log\frac{\prod_{k\geq 1}\left(1-\frac{1}{(2k)^2 a^2}\right)}{\prod_{k\geq 0}\left(1-\frac{1}{(2k+1)^2 a^2}\right)}\tag{3} $$ y el reclamo de la siguiente manera a partir de la Weierstrass producto para el seno y el coseno funciones.