A generalized log sine integral .

In summary, the conversation discusses finding a general formula for the integral I(a,t) = \int^t_0 x \log|\sin(a x )| \, dx. The conversation also delves into a special case of the integral and discusses potential substitutions and series expansions to simplify the formula.
  • #1
alyafey22
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This thread will be dedicated to find a general formula for the integral
\(\displaystyle I(a,t) = \int^t_0 x \log|\sin(a x )| \, dx \,\,\,\,\, a,t>0\)​

This is not a tutorial . Any comments or attempts are always be welcomed .
 
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  • #2
Re: A generalized log gamma integral .

We consider the special case

\(\displaystyle I\left(1,\frac{\pi}{2} \right) = \int^{\frac{\pi}{2}}_0 x \log|\sin(x )| \, dx =\int^{\frac{\pi}{2}}_0 x \log|2\sin(x )| -\frac{\pi^2}{8}\log(2)\)

Now we integrate by parts

\(\displaystyle
\begin{align}
\int^{\frac{\pi}{2}}_0 x \log|2\sin(x )| dx &=\frac{1}{2} \int^{\frac{\pi}{2}}_0 \mathrm{Cl}_2(2\theta)\, d\theta\\ &=\frac{1}{2} \int^{\frac{\pi}{2}}_0 \sum_{n=1}^{\infty}\frac{\sin(2n\theta)}{n^2}\, d\theta \\ &=-\frac{1}{4}\sum_{n=1}\frac{(-1)^n}{n^3}+\frac{1}{4}\sum_{n=1}\frac{1}{n^3}\\
&=\frac{7}{16}\zeta(3)
\end{align}
\)

Eventually we have

\(\displaystyle I\left(1,\frac{\pi}{2} \right) = \frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\log(2)\)
 
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  • #3
Re: A generalized log gamma integral .

For the generalized form we need to find the general integral

\(\displaystyle I_0(t,1) = \int^t_0 \mathrm{Cl}_2(\theta) \, d\theta \)

It is easy to see that

\(\displaystyle \mathrm{Cl}(\theta) = \Im \left(\mathrm{Li}_2(e^{i\theta}) \right)\)

We already proved in separately that

\(\displaystyle \Im \left(\mathrm{Li}_2(e^{i\theta}) \right) = \frac{\mathrm{Li}_2(e^{i\theta}) -\mathrm{Li}_2(e^{-i\theta}) }{2i}\)

Now consider

\(\displaystyle \int^t_0 \mathrm{Li}_2(e^{i\theta})\, d\theta \)

Now we consider $t\in (0,2\pi]$ and let $z=e^{i\theta }$ hence $-i \log(z) = \theta $

\(\displaystyle -i\int^{e^{i\theta}}_{1} \frac{\mathrm{Li}_2(z)}{z} \, dz = -i \left(\mathrm{Li}_3(e^{i\theta}) -\zeta(3)\right) \)

Simalrily we have

\(\displaystyle \int^t_0 \mathrm{Li}_2(e^{-i\theta})\, d\theta =\int^{e^{-i\theta}}_{1} \frac{\mathrm{Li}_2(z)}{z} \, dz = i \left(\mathrm{Li}_3(e^{-i\theta}) -\zeta(3)\right)\)

Hence we have

\(\displaystyle \frac{-i \left(\mathrm{Li}_3(e^{i\theta}) -\zeta(3)\right)-i \left(\mathrm{Li}_3(e^{-i\theta}) -\zeta(3)\right)}{2i}=-\frac{\mathrm{Li}_3(e^{i\theta})+\mathrm{Li}_3(e^{-i\theta})}{2}+\zeta(3)\)

I conjucture that

\(\displaystyle I_0(t,1) = -\Re \left( \mathrm{Li}_3 (e^{it}) \right)+\zeta(3)\)

For the special case

\(\displaystyle I_0 \left(\frac{\pi}{2},1 \right)= \frac{7}{4} \zeta(3)\)

Gotta rush now , I hope I didn't make mistakes :) .
 
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  • #4
\begin{align}
I(a,t) = \int^t_0 x \log|\sin(a x )| \, dx &= \frac{1}{4a^2} \int^{2at}_0 x \log |\sin \frac{x}{2}|\, dx \\ &= \frac{1}{4a^2} \int^{2at}_0 x \log |2\sin \frac{x}{2}|\, dx-\frac{\log(2)}{2} t^2\\ &= -\frac{t}{2a}\mathrm{Cl}_2(2at) + \frac{1}{4a^2}\int^{2at}_0\mathrm{Cl}_2(x) dx-\frac{\log(2)}{2} t^2\\
&= -\frac{t}{2a}\mathrm{Cl}_2(2at)+\frac{1}{4a^2} \left( \zeta(3)-\Re \left( \mathrm{Li}_3 (e^{2i \, at}) \right) \right)-\frac{\log(2)}{2} t^2
\end{align}

Hence we have

\(\displaystyle I(a,t) = -\frac{t}{2a}\mathrm{Cl}_2(2at)+\frac{1}{4a^2} \left( \zeta(3)-\Re \left( \mathrm{Li}_3 (e^{2i \, at}) \right) \right)-\frac{\log(2)}{2} t^2 \)

Hence we have for $a=\frac{1}{2}$

\(\displaystyle I \left(\frac{1}{2},t \right) = -t\, \mathrm{Cl}_2(t)-\Re \left( \mathrm{Li}_3\, e^{i \, t} \right)-\frac{\log(2)}{2} t^2+\zeta(3)\)

or we have

\(\displaystyle I \left(\frac{1}{2},t \right) = -t\, \mathrm{Cl}_2(t)- \mathrm{Cl}_3(t)-\frac{\log(2)}{2} t^2+\zeta(3)\)
 
  • #5
Here's a little something you might find interesting, Zaid... ;)Let's say you evaluate the function

\(\displaystyle I(a,t) = \int^t_0 x \log|\sin(a x )| \, dx \)

for a few particular values of the parameters \(\displaystyle a\) and \(\displaystyle t\), in terms of Clausen functions, etc. Next, perform the substitution \(\displaystyle y=ax\) to obtain\(\displaystyle I(a,t) = \frac{1}{a^2}\int^{at}_0 y \log|\sin y| \, dx \)After that, provided that \(\displaystyle 0 < at < \pi\) - whereby you can also drop the absolute value sign in the integrand - you can apply the logsine series result:\(\displaystyle \log (\sin x) = \log x + \sum_{k=1}^{\infty}(-1)^k\frac{2^{2k-1} B_{2k}}{k (2k)!} x^{2k} \quad [ \text{valid for} \, 0 < x < \pi]\)to get\(\displaystyle I(a,t) = \frac{1}{a^2}\int^{at}_0 x\log x \,dx + \frac{1}{a^2} \sum_{k=1}^{\infty}(-1)^k\frac{2^{2k-1} B_{2k}}{k (2k)!} \int_0^{at} x^{2k+1}\,dx=\)\(\displaystyle \frac{t^2}{4}(2\log (at)-1) + \frac{1}{a^2} \sum_{k=1}^{\infty}(-1)^k\frac{2^{2k-1} B_{2k}}{k (2k)!} \int_0^{at} x^{2k+1}\,dx=\)\(\displaystyle \frac{t^2}{4}(2\log (at)-1) + \frac{1}{a^2} \sum_{k=1}^{\infty}(-1)^k\frac{2^{2k-1} B_{2k}}{k (2k)! (2k+2) } (at)^{2k+2} \)
Next, take the classic Zeta function result\(\displaystyle \zeta(2k)=(-1)^{k+1} \frac{(2\pi)^{2k} B_{2k}}{2(2k)!} \quad k \in \mathbb{Z} \ge 1\)and invert the terms to express the Bernoulli numbers as\(\displaystyle B_{2k}=2(-1)^{k+1}\frac{(2k)!}{(2\pi)^{2k}} \zeta(2k)\)Substitute this back into the series result to obtain\(\displaystyle I(a,t) = \frac{t^2}{4}(2\log (at)-1) +\)

\(\displaystyle \frac{1}{a^2} \sum_{k=1}^{\infty}(-1)^k\frac{2^{2k-1} }{k (2k)! (2k+2) } \left[ 2(-1)^{k+1}\frac{(2k)!}{(2\pi)^{2k}} \zeta(2k) \right] (at)^{2k+2} =\)\(\displaystyle \frac{t^2}{4}(2\log (at)-1) - \frac{1}{2a^2} \sum_{k=1}^{\infty} \frac{ \zeta(2k) }{k(k+1)\pi^{2k}} (at)^{2k+2}\)Finally, use the explicit evaluations you have of the function \(\displaystyle I(a,t)\) - provided that \(\displaystyle 0 < at < \pi\) - and you have a closed form evaluation for the Zeta Series above:
\(\displaystyle \sum_{k=1}^{\infty} \frac{ \zeta(2k) }{k(k+1)\pi^{2k}} (at)^{2k+2} = \frac{a^2 t^2}{2}(2\log (at)-1) - 2a^2 \, I(a,t)\)

(Heidy)(Heidy)(Heidy)NB. Made a bit of a typo in there to start with, but hopefully it's all fixed now... Main thing is the process, anyhoo lol
 
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  • #6
Well, using the Lewin's book using entries [4.18],[6.18] and [16.23] , I got the following

\(\displaystyle I\left(\frac{1}{2},\frac{\pi}{3} \right)=\int^{\frac{\pi}{3}}_0 \theta \, \log \left( \sin \frac{\theta}{2} \right)\, d\theta = -\frac{\pi}{3}\mathrm{Cl}_2\left( \frac{\pi}{3}\right)-\frac{\pi^2 \log(2)}{18}+\frac{2}{3}\zeta(3)\)

\(\displaystyle I\left(\frac{1}{2},\frac{2\pi}{3} \right)=\int^{\frac{2\pi}{3}}_0 \theta \, \log \left( \sin \frac{\theta}{2} \right)\, d\theta = -\frac{4\pi}{9}\mathrm{Cl}_2\left( \frac{\pi}{3}\right)-\frac{2\pi^2 \log(2)}{9}+\frac{13}{9}\zeta(3)\)

By some manipulations we have

\(\displaystyle \int^{\frac{2\pi}{3}}_0 \theta \, \log \left( \sin \frac{\theta}{2} \right)\, d\theta =4\int^{\frac{\pi}{3}}_0 \theta \, \log \left( \sin \theta \right)\, d\theta =4\int^{\frac{\pi}{3}}_0 \theta \log(2) \, d\theta +4\int^{\frac{\pi}{3}}_0 \theta \, \log \left( \sin\frac{\theta}{2} \right)\, d\theta+4\int^{\frac{\pi}{3}}_0 \theta \, \log \left(\cos \frac{\theta}{2} \right)\, d\theta \)

Hence we have

\(\displaystyle \int^{\frac{\pi}{3}}_0 \theta \, \log \left(\cos \frac{\theta}{2} \right)\, d\theta = -\frac{\pi^2 \log(2)}{18}+\frac{1}{4} I \left( \frac{1}{2},\frac{2\pi}{3}\right)-I\left( \frac{1}{2},\frac{\pi}{3}\right) \)

A simplification could be done , finish it later .
 
  • #7
DreamWeaver said:
\(\displaystyle \sum_{k=1}^{\infty} \frac{ \zeta(2k) }{k(k+1)\pi^{2k}} (at)^{2k+2} = \frac{a^2 t^2}{2}(2\log (at)-1) - 2a^2 \, I(a,t)\)

Woow DW , very nice ! I liked it .
 
  • #8
So you're not that 'trig-shy' after all, Zaid... Good stuff! :D:D:D
 
  • #9
we conclude this thread by pointing out the results we have \(\displaystyle \tag{1} \, \int^{\frac{\pi}{3}}_0 \theta \, \log \left(\cos \frac{\theta}{2} \right)\, d\theta = \frac{2\pi}{9}\mathrm{Cl}_2\left( \frac{\pi}{3}\right)-\frac{\pi^2 \log(2)}{18} -\frac{11}{36}\zeta(3)\)

\(\displaystyle \tag{2} \int^{\frac{\pi}{3}}_0 \theta \, \log \left( \sin \frac{\theta}{2} \right)\, d\theta = -\frac{\pi}{3}\mathrm{Cl}_2\left( \frac{\pi}{3}\right)-\frac{\pi^2 \log(2)}{18}+\frac{2}{3}\zeta(3)\)

\(\displaystyle \tag{3} \int^{\frac{2\pi}{3}}_0 \theta \, \log \left( \sin \frac{\theta}{2} \right)\, d\theta = -\frac{4\pi}{9}\mathrm{Cl}_2\left( \frac{\pi}{3}\right)-\frac{2\pi^2 \log(2)}{9}+\frac{13}{9}\zeta(3)\)

It seems that we cannot represent \(\displaystyle \mathrm{Cl}_2\left( \frac{\pi}{3}\right)\) in terms of elementary functions. We could get more results by exploring the integrals with argument \(\displaystyle \frac{\pi}{2}\). I think we shall not consider that because they can be derived easily. Ok that is it and we conclude this thread.
 

Related to A generalized log sine integral .

1. What is a generalized log sine integral?

A generalized log sine integral is a mathematical function that is used to solve problems in analysis, number theory, and physics. It is defined as the integral of the logarithm of the sine function, and is denoted by Lin(x).

2. How is a generalized log sine integral calculated?

The generalized log sine integral can be calculated using various methods, such as numerical integration or series expansion. It is also related to other special functions, such as the polylogarithm function and the Riemann zeta function.

3. What are the main properties of the generalized log sine integral?

The generalized log sine integral has several important properties, including the functional equation Lin(x) = (-1)n+1Lin(1/x) and the recurrence relation Lin+1(x) = -ln(1-x) - Lin(x). It also has asymptotic behavior for large and small values of x.

4. In what areas of science is the generalized log sine integral used?

The generalized log sine integral has applications in many areas of science, including physics, number theory, and complex analysis. It is used to solve problems related to the Riemann zeta function, the distribution of prime numbers, and the behavior of quantum systems.

5. Are there any real-life applications of the generalized log sine integral?

Yes, the generalized log sine integral has practical applications in fields such as electrical engineering and signal processing. It is used to model the behavior of electrical circuits and to analyze signals in communication systems. It is also used in the design of filters and oscillators.

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