11
$\begingroup$

Consider the integral $(1)$

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

An attempt:

Rewrite $(1)$ as

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

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

Or we can rewrite $(1)$ as

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

Applying $\ln\tan x$ series to the red part

$I_1$ becomes

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

This looked too complicate, how else can we prove $(1)?$

$\endgroup$
  • 1
    $\begingroup$ If $x>\pi/4, \tan x>1$ right? Is it a problem on Complex calculus? $\endgroup$ – lab bhattacharjee Apr 5 '17 at 6:22
  • 1
    $\begingroup$ @lab Oh thats why I was getting $\ln(-1)$ when solving this. $\endgroup$ – samjoe Apr 5 '17 at 6:23
  • $\begingroup$ have you tried $x\rightarrow x-\pi/2$? $\endgroup$ – tired Apr 5 '17 at 6:59
  • $\begingroup$ additionally @labbhattacharjee is right: something is fishy here $\endgroup$ – tired Apr 5 '17 at 7:02
  • 1
    $\begingroup$ If you change the interval of integration for $[0;\tfrac{\pi}{4}]$ the result is probably $\dfrac{13}{72}$ $\endgroup$ – FDP Apr 5 '17 at 9:16
11
+50
$\begingroup$

What you need is to split the integral into two parts and use integration by parts for the second part. Let $x\to\frac{\pi}{2}-x$ and then \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)+\pi 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} Here $$ \int^{\pi/4}_{0}(\frac{\pi}{2}-x)\cos(8x)\mathrm dx=0. $$

$\endgroup$
2
$\begingroup$

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

I'll assume the $\ds{\ln}$-argument is 'enclosed' in an absolute value !!!.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.