Cómo podemos mostrar que $\int_{0}^{\pi/2}x\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx={\pi\over 12}?$

Question

Considere el integral $(1)$

$$\int_{0}^{\pi/2}x\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx={\pi\over 12}\tag1$$

Un intento:

Reescritura de $(1)$ como

$$\int_{0}^{\pi/2}x\cos(8x)\ln\tan\left(x+{\pi\over 4}\right)\mathrm dx\tag2$$

$$\int_{0}^{\pi/2}x\cdot{1-\tan^2(4x)\over 1+\tan^2(4x)}\ln\tan\left(x+{\pi\over 4}\right)\mathrm dx\tag3$$

O podemos reescribir $(1)$ como

$$\color{red}{\int_{0}^{\pi/2}x\cos^2(4x)\ln\tan\left(x+{\pi\over 4}\right)\mathrm dx}-\int_{0}^{\pi/2}x\sin^2(4x)\ln\tan\left(x+{\pi\over 4}\right)\mathrm dx=I_1+I_2\tag4$$

Aplicando serie de #% de %#% a la parte roja

$\ln\tan x$ se convierte

$I_1$$

Esto parecía demasiado complicado, cómo podemos demostrar $$I_1=\int_{0}^{\pi/2}x\cos^2(4x)\ln\left(x+{\pi\over 4}\right)\mathrm dx+\sum_{n=1}^{\infty}{2^{2n}(2^{2n-1}-1)B_n\over n(2n)!}\int_{0}^{\pi/2}x(x+\pi/4)^{2n}\cos^2(4x)\mathrm dx\tag5$

Answer 1

Lo que necesitas es dividir la integral en dos partes y usar integración por partes. Que $x\to\frac{\pi}{2}-x$ y luego\begin{eqnarray} I&=:&\int_{0}^{\pi/2}x\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx\\ &=&\int_{0}^{\pi/4}x\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx+\int_{\pi/4}^{\pi/2}x\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx\\ &=&\int_{0}^{\pi/4}x\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx-\int_{\pi/4}^{0}(\frac{\pi}{2}-x)\cos(8x)\ln\left(1+\cot x\over 1-\cot x\right)\mathrm dx\\ &=&\int_{0}^{\pi/4}x\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx+\int^{\pi/4}_{0}(\frac{\pi}{2}-x)\cos(8x)\ln\left(1+\tan x\over \tan x-1\right)\mathrm dx\\ &=&\int_{0}^{\pi/4}x\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx+\int^{\pi/4}_{0}(\frac{\pi}{2}-x)\cos(8x)\left[\ln\left(1+\tan x\over 1-\tan x\right)+i\right]\mathrm dx\\ &=&\int_{0}^{\pi/4}x\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx+\int^{\pi/4}_{0}(\frac{\pi}{2}-x)\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx\\ &&+\pi i\int^{\pi/4}_{0}(\frac{\pi}{2}-x)\cos(8x)\mathrm dx\\ &=&\frac{\pi}{2}\int^{\pi/4}_{0}\cos(8x)\ln\left(1+\tan x\over 1-\tan x\right)\mathrm dx\\ &=&\frac{\pi}{16}\int^{\pi/4}_{0}\ln\left(1+\tan x\over 1-\tan x\right)\mathrm d\sin(8x)\\ &=&\frac{\pi}{16}\left[\ln\left(1+\tan x\over 1-\tan x\right)\sin(8x)\bigg|_0^{\pi/4}-\int^{\pi/4}_{0}\sin(8x)\mathrm{d}\ln\left(1+\tan x\over 1-\tan x\right)\right]\\ &=&-\frac{\pi}{16}\int^{\pi/4}_{0}\sin(8x)\left(\frac1{1+\tan x}+\frac1{ 1-\tan x}\right)\sec^2x\mathrm d x\\ &=&-\frac{\pi}{8}\int^{\pi/4}_{0}\frac{\sin(8x)}{\cos(2x)}\mathrm d x\\ &=&\frac{\pi}{12}. \end{eqnarray} es fácil comprobar %#% $ #%

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Voy a asumir el $\ds{\ln}$-argumento es 'encerrado' en un valor absoluto !!!.

\begin{align} &\int_{0}^{\pi/2}x\cos\pars{8x} \ln\pars{\verts{1 + \tan\pars{x} \over 1 - \tan\pars{x}}}\,\dd x = \int_{0}^{\pi/2}x\cos\pars{8x} \ln\pars{\verts{\tan\pars{x + {\pi \over 4}}}}\,\dd x \\[5mm] = &\ -\int_{-\pi/4}^{\pi/4}\pars{x + {\pi \over 4}}\cos\pars{8x} \ln\pars{\verts{\tan\pars{x}}}\,\dd x = -\,{\pi \over 2}\int_{0}^{\pi/4}\cos\pars{8x} \ln\pars{\verts{\tan\pars{x}}}\,\dd x \\[5mm] = &\ {\pi \over 16}\int_{0}^{\pi/4}\sin\pars{8x} {\sec^{2}\pars{x} \over \tan\pars{x}}\,\dd x = {\pi \over 8}\int_{0}^{\pi/4}{\sin\pars{8x} \over \sin\pars{2x}}\,\dd x = {\pi \over 16}\,\Im\int_{0}^{\pi/2}{\expo{4\ic x} \over \sin\pars{x}}\,\dd x \\[5mm] = &\ \left.{\pi \over 16}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2} {z^{4} \over \pars{1 - z^{2}}\ic/\pars{2z}} \,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} = \left.-\,{\pi \over 8}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}{z^{4} \over 1 - z^{2}} \,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}&\ {\pi \over 8}\,\Im\int_{1}^{0}{y^{4} \over 1 + y^{2}}\,\ic\,\dd y\ +\ \underbrace{% {\pi \over 8}\,\Im\int_{0}^{1 - \epsilon}{x^{4} \over 1 - x^{2}}\,\dd x} _{\ds{=\ 0}}\ +\ {\pi \over 8}\,\Im\int_{\pi}^{\pi/2}{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \pars{-\epsilon\expo{\ic\theta}}\pars{2}} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to} & -\,\ \underbrace{{\pi \over 8}\int_{0}^{1}{y^{4} \over y^{2} + 1}\,\dd y} _{\ds{{\pi^{2} \over 32} - {\pi \over 12}}} + {\pi^{2} \over 32} = \bbx{\ds{\pi \over 12}} \end{align}