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Log-trig integral

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Post Fri Aug 19, 2016 10:18 am
FDP

Posts: 22
$\displaystyle \int_0^{\tfrac{\pi}{4}} \ln(\tan x)\ln(\cos x-\sin x)dx=\dfrac{G\ln 2}{2}$

Don't expect an answer from me, i don't know, for now, how to compute it.

Post Fri Aug 19, 2016 2:07 pm

Posts: 47

FDP wrote:
$\displaystyle \int_0^{\tfrac{\pi}{4}} \ln(\tan x)\ln(\cos x-\sin x)dx=\dfrac{G\ln 2}{2}$

Don't expect an answer from me, i don't know, for now, how to compute it.


\begin{align*}
\int_{0}^{\pi/2} \log \left (\tan x \right ) \log \left ( \cos x - \sin x \right ) \, {\rm d}x &= \int_{0}^{\pi/4} \log \left (\tan x \right ) \log \left [ \sqrt{2} \sin \left (\frac{\pi}{4}-x \right ) \right ] \, {\rm d}x \\
&= \int_0^{\pi/4} \log \left ( \tan x \right ) \log \sqrt{2} \, {\rm d}x + \int_{0}^{\pi/4} \log (\tan x) \log \left [ \sin \left ( \frac{\pi}{4}-x \right ) \right ] \, {\rm d}x \\
&= \frac{1}{2} \log 2 \int_{0}^{\pi/4} \log \left ( \tan x \right ) \, {\rm d}x + \int_{0}^{\pi/4} \log (\tan x) \log \left [ \sin \left ( \frac{\pi}{4}-x \right ) \right ] \, {\rm d}x \\
&= - \frac{\mathcal{G} \ln 2}{2} + \int_{0}^{\pi/4} \log \left [ \tan \left ( \frac{\pi}{4}-x \right ) \right ] \log \sin x \, {\rm d}x \\
&=- \frac{\mathcal{G} \ln 2}{2} + \mathcal{G} \ln 2 \\
&= \frac{\mathcal{G} \ln 2}{2}
\end{align*}

$(*)$ For the evaluation of the $\log \tan $ integral we use the fact that $\tan x =\frac{\cos x}{\sin x}$ and then the Fourier series of $\log \sin$ and $\log \cos$ respectively. Now for the remaining integral I have no idea how to tackle it. I am talking of course about

$$\int_{0}^{\pi/4} \log \left [ \tan \left ( \frac{\pi}{4}-x \right ) \right ] \log \sin x \, {\rm d}x$$

Based on your result I was able to actually conjecture its closed form. Prove it however, seems a bit of a challenge. Any help?

Edit (a few minutes later)

\begin{align*}
\int_{0}^{\pi/4} \log \left [ \tan \left ( \frac{\pi}{4}-x \right ) \right ]\log (\sin x) \, {\rm d}x&=\int_{0}^{\pi/4} \log \left [ \frac{\tan \frac{\pi}{4} - \tan x}{1+\tan \frac{\pi}{4} \tan x} \right ] \log \left ( \sin x \right ) \, {\rm d}x \\
&= \int_{0}^{\pi/4} \log \left [ \frac{1-\tan x}{1+\tan x} \right ] \log (\sin x) \, {\rm d}x\\
&= \int_{0}^{\pi/4} \log \left [ \frac{\left ( 1-\tan x \right )^2}{1-\tan^2 x}\right ] \log \left (\sin x \right ) \, {\rm d}x\\
&= 2\int_{0}^{\pi/4} \log \left ( 1-\tan x \right ) \log \left ( \sin x \right ) \, {\rm d}x - \int_{0}^{\pi/4}\log \left ( 1-\tan^2 x \right ) \log \left ( \sin x \right ) \, {\rm d}x \\
&= \int_{0}^{\pi/4} \log \left ( 1-\tan x \right ) \log \left ( \sin x \right ) \, {\rm d}x - \int_{0}^{\pi/4} \log \left ( 1+\tan x \right ) \log (\sin x) \, {\rm d}x
\end{align*}

This seems like a good start. Anyone to finish it off, as I am not sure how to progress?


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