Una manera diferente para demostrar que $\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \frac{3 G}{4} + \frac{\pi}{16} \, \ln 2$

Question

El hecho de que $$\int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy $$ is an integral representation of Catalan's constant ($G$), I was able to deduce that $$\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \int_{0}^{\pi/8} \ln(1+ \cot u) \, du = \frac{3G}{4} + \frac{\pi}{16} \, \ln 2\tag{1}.$$

Lo que es otra forma de probar la $(1)$ que de preferencia no implica la dilogarithm función?

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EDITAR:

En respuesta al Dr. MV del comentario, el siguiente es cómo deduje $(1)$ a partir de esa representación integral de catalán constante.

$$ \begin{align} G&= \int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{2-y^{2}} \int_{0}^{1} \frac{1}{1-\frac{x^{2}}{2-y^{2}}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{\sqrt{2-y^{2}}} \, \text{artanh} \left(\frac{1}{\sqrt{2-y^{2}}} \right) \, dy \\ &= \int_{0}^{\pi/4} \text{artanh} \left(\frac{1}{\sqrt{2} \cos \theta} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{\sqrt{2} \cos \theta +1}{\sqrt{2} \cos \theta -1} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{(\sqrt{2} \cos \theta+1)^{2}}{2 \cos^{2} \theta -1} \right) \, d \theta \\ &=\int_{0}^{\pi/4} \ln (\sqrt{2} \cos \theta +1) \, d \theta - \frac{1}{2} \int_{0}^{\pi/4} \ln(\cos 2 \theta) \, d \theta \\ &= \int_{0}^{\pi/4} \ln\left(\sqrt{2} \cos \left(\frac{\pi}{4} - \phi\right)+1\right) \, d \phi - \frac{1}{4} \int_{0}^{\pi/2} \ln( \cos \tau) \, d \tau \\ &= \int_{0}^{\pi/4} \ln \left(\sin(\phi) + \cos(\phi)+1\right) \, d \phi - \frac{1}{4} \left(- \frac{\pi}{2} \, \ln 2 \right) \\ &= \int_{0}^{\pi/4} \ln (\sin \phi) \, d \phi + \int_{0}^{\pi/4} \ln \left(1+ \frac{1+ \cos \phi}{\sin \phi} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{4} \, \ln 2 + \int_{0}^{\pi/4} \ln \left(1+ \cot \frac{\phi}{2} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{8} \, \ln 2 + 2 \int_{0}^{\pi/8} \ln (1 + \cot u) \, du \end{align}$$

Answer 1

Mediante el siguiente trigonométricas identitiy:

$$(1+\cot x)^2 =\frac{ 2 \tan\left(\frac{\pi}{4}+x\right)}{\tan(2 x) \, \tan(x)},$$ Tenemos \begin{align}2 \int_0^{\pi/8} \ln(1+ \cot x) \, dx= & \ln2 \int_0^{\pi/8} dx-\int_0^{\pi/8} \ln \tan 2 x \, dx \\ - & \int_0^{\pi/8} \ln \tan x \, dx+\int_0^{\pi/8} \ln \tan\left(\frac{\pi}{4}+x\right)\, dx \tag{1} \\ = & \frac{\pi}{8} \ln 2-\frac12 \int_0^{\pi/4} \ln \tan x \, dx \\ - & \int_0^{\pi/8} \ln \tan x \, dx-\int_{\pi/8}^{\pi/4} \ln \tan x \, dx \tag{2} \\ = & \frac{\pi}{8} \ln 2-\frac32 \int_0^{\pi/4} \ln \tan x \, dx \\= & \frac{\pi}{8} \ln 2 + \frac32 G. \end{align}

Donde, para llegar a $(2)$ $(1)$ hemos sustituido $x \mapsto x/2$ en la segunda integral, y en la última integral nosotros sustituido $x \mapsto \pi/4-x$, en tándem con el hecho de que $\tan(\pi/2-x) = 1/\tan(x)$.

También, la integral de la $\int_0^{\pi/4} \ln \tan x \, dx = - G$ es bien conocida y puede ser fácilmente reproducido.

Answer 2

El uso de la identidad

$$ \log(\cos(x))=-\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\cos(2kx)}k\etiqueta{1} $$

y las evaluaciones

$$ \begin{align} \sin(3k\pi/4)-\sin(k\pi/4) &=2\color{#C00000}{\cos(k\pi/2)}\color{#00A000}{\sin(k\pi/4)}\\ &=\left\{\begin{array}{} \color{#C00000}{0}&\text{if %#%#% is odd}\\ 2(-1)^{\frac{k/2+1}2}&\text{if %#%#% is odd}\\ \color{#00A000}{0}&\text{if %#%#% is even} \end{array}\right.\la etiqueta{2}\\ \sin(k\pi/2)&=\left\{\begin{array}{} (-1)^{\frac{k-1}2}&\text{if %#%#% is odd}\\ 0&\text{if %#%#% is even} \end{array}\right.\la etiqueta{3} \end{align} $$

tenemos $$ \begin{align} &\int_{1+\sqrt2}^\infty\frac{\log(1+x)}{1+x^2}\,\mathrm{d}x\tag{4a}\\ &=\int_{3\pi/8}^{\pi/2}\log(1+\tan(x))\,\mathrm{d}x\tag{4b}\\ &=\int_{3\pi/8}^{\pi/2}\left[\log(\cos(x)+\sin(x))-\log(\cos(x))\right]\,\mathrm{d}x\tag{4c}\\ &=\int_{3\pi/8}^{\pi/2}\left[\log(\sqrt2\cos(x-\pi/4))-\log(\cos(x))\right]\,\mathrm{d}x\tag{4d}\\ &=\frac\pi{16}\log(2)+\int_{\pi/8}^{\pi/4}\log(\cos(x))\,\mathrm{d}x-\int_{3\pi/8}^{\pi/2}\log(\cos(x))\,\mathrm{d}x\tag{4e}\\ &=\frac\pi{16}\log(2)+\left[\sum_{k=1}^\infty(-1)^{k-1}\frac{\sin(2kx)}{2k^2}\right]_{\pi/8}^{\pi/4}\,-\left[\sum_{k=1}^\infty(-1)^{k-1}\frac{\sin(2kx)}{2k^2}\right]_{3\pi/8}^{\pi/2}\tag{4f}\\ &=\frac\pi{16}\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\sin(k\pi/2)-\sin(k\pi/4)+\sin(3k\pi/4)}{2k^2}\tag{4g}\\ &=\frac\pi{16}\log(2)+\sum_{k=0}^\infty\frac{(-1)^k}{2(2k+1)^2}+\frac{2(-1)^k}{2(4k+2)^2}\tag{4h}\\ &=\frac\pi{16}\log(2)+\frac34\mathrm{G}\tag{4i} \end{align} $$ Explicación:
$k$: sustituto $k/2$
$k/2$: $k$
$k$: $\text{(4b)}$
$x\mapsto\tan(x)$: sustituto $\text{(4c)}$ en la izquierda integral
$(1+\tan(x))\cos(x)=\cos(x)+\sin(x)$: aplicar $\text{(4d)}$
$\cos(x)+\sin(x)=\sqrt2\cos(x-\pi/4)$: evaluar en los límites
$\text{(4e)}$: aplicar $x\mapsto x+\pi/4$ $\text{(4f)}$
$(1)$: aplicar la definición de catalán Constante

Answer 3

Realizar el cambio de variable $y=\arctan x$,

$\displaystyle I=\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln(1+\tan x)dx=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}} \ln\left(\sin x+\cos x\right)dx-\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos x\right)dx$

$\begin{align}\displaystyle \int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}} \ln\left(\sin x+\cos x\right)dx&=\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\sqrt{2}\cos\left(\dfrac{\pi}{4}-x\right)\right)dx\\ &=\dfrac{\pi}{16}\ln 2+\int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos\left(\dfrac{\pi}{4}-x\right)\right)dx\\ &=\dfrac{\pi}{16}\ln 2+\int_{\tfrac{\pi}{8}}^{\tfrac{\pi}{4}}\ln(\cos x)dx\\ &=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln(\cos x)dx \end{align}$

Realizar el cambio de variable $y=\dfrac{\pi}{2}-x$,

$\displaystyle \int_{\tfrac{3\pi}{8}}^{\tfrac{\pi}{2}}\ln\left(\cos x\right)dx=\int_{0}^{\tfrac{\pi}{8}}\ln\left(\sin x\right)dx$

Por lo tanto,

$\begin{align} \displaystyle I&=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln\left(\sin x\right)dx\\ &=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln(\cos x\sin x)dx\\ &=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_{0}^{\tfrac{\pi}{8}}\ln\left(\dfrac{\sin(2x)}{2}\right)dx\\ &=\dfrac{\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln\left(\dfrac{\sin x}{2}\right)dx\\ &=\dfrac{3\pi}{16}\ln 2+\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx-\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln\left(\sin x\right)dx\\ &=\dfrac{3\pi}{16}\ln 2+\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx+\dfrac{1}{2}\int_{0}^{\tfrac{\pi}{4}}\ln(\cot x)dx \end{align}$

Desde entonces,

$\displaystyle \int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx=\dfrac{1}{2} G-\dfrac{1}{4}\pi\ln 2$

y,

$\displaystyle \int_{0}^{\tfrac{\pi}{4}}\ln(\cot x)dx=G$

a continuación,

$\begin{align} I&=\dfrac{3\pi}{16}\ln 2+\dfrac{1}{4} G-\dfrac{1}{8}\pi\ln 2+\dfrac{1}{2}G\\ &=\dfrac{\pi}{16}\ln 2+\dfrac{3}{4}G \end{align}$

PS:

Prueba de

$\displaystyle \int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx=\dfrac{1}{2} G-\dfrac{1}{4}\pi\ln 2$

$\begin{align} \displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx&=\int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}}\ln(\sin x)dx\\ &=\int_0^{\tfrac{\pi}{2}}\ln(\sin x)dx\\ &=2\int_0^{\tfrac{\pi}{4}}\ln\left(\sin(2x)\right)dx\\ &=2\int_0^{\tfrac{\pi}{4}}\ln\left(2\sin x\cos x\right)dx\\ &=\dfrac{1}{2}\pi\ln 2+2\int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+2\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx \end{align}$

Por lo tanto,

$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx+\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx=-\dfrac{1}{2}\pi\ln 2$

y,

$\displaystyle G=\int_0^{\tfrac{\pi}{4}}\ln(\cot x)dx=\int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx-\int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx$

por lo tanto,

$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\cos x)dx=\dfrac{1}{2}G-\dfrac{1}{4}\pi\ln 2$

$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\sin x)dx=-\dfrac{1}{2}G-\dfrac{1}{4}\pi\ln 2$