MHB A generalized log sine integral .

[alyafey22]

This thread will be dedicated to find a general formula for the integral

$$I(a,t) = \int^t_0 x \log|\sin(a x )| \, dx \,\,\,\,\, a,t>0$$​

This is not a tutorial . Any comments or attempts are always be welcomed .

 

[alyafey22]

Re: A generalized log gamma integral .

We consider the special case

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

Now we integrate by parts

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

Eventually we have

$$I\left(1,\frac{\pi}{2} \right) = \frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\log(2)$$

 

[alyafey22]

Re: A generalized log gamma integral .

For the generalized form we need to find the general integral

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

It is easy to see that

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

We already proved in separately that

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

Now consider

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

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

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

Simalrily we have

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

Hence we have

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

I conjucture that

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

For the special case

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

Gotta rush now , I hope I didn't make mistakes :) .

 

[alyafey22]

\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

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

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

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

or we have

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

 

[DreamWeaver]

Here's a little something you might find interesting, Zaid... ;)Let's say you evaluate the function

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

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

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

(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

 

[alyafey22]

Well, using the Lewin's book using entries [4.18],[6.18] and [16.23] , I got the following

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

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

By some manipulations we have

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

Hence we have

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

A simplification could be done , finish it later .

 

[alyafey22]

$$\sum_{k=1}^{\infty} \frac{ \zeta(2k) }{k(k+1)\pi^{2k}} (at)^{2k+2} = \frac{a^2 t^2}{2}(2\log (at)-1) - 2a^2 \, I(a,t)$$

Woow DW , very nice ! I liked it .

 

[DreamWeaver]

So you're not that 'trig-shy' after all, Zaid... Good stuff! :D:D:D

 

[alyafey22]

we conclude this thread by pointing out the results we have $$\tag{1} \, \int^{\frac{\pi}{3}}_0 \theta \, \log \left(\cos \frac{\theta}{2} \right)\, d\theta = \frac{2\pi}{9}\mathrm{Cl}_2\left( \frac{\pi}{3}\right)-\frac{\pi^2 \log(2)}{18} -\frac{11}{36}\zeta(3)$$

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

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

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