5
$\begingroup$

These are known as LogSine integrals at $2\pi/3$, so I will call the integral Ls as this is common in the literature. I am trying to prove $$ Ls=-\int_0^{\pi/3} \ln^2\big(2\cos \frac{\theta}{2}\big) d\theta=-\frac{13\pi^3}{162}-2Gl_{2,1}\big(\frac{2\pi}{3}\big) $$ where $Gl_{2,1}$ can be reduced to one-dimensional polylogarithmic constants. I know we can write $$ \ln^2\big(2\cos \frac{\theta}{2}\big) =\big(\ln 2+\ln \cos\frac{\theta}{2}\big)^2=\ln^2 2+\ln^2 \cos \frac{\theta}{2} +2\ln 2 \ln \cos \frac{\theta}{2}, $$ but am totally stuck at this point. Thanks

$\endgroup$
  • $\begingroup$ What's the closed form? $\endgroup$ – Antonio Vargas Apr 12 '14 at 23:58
  • $\begingroup$ @AntonioVargas i have added the closed form for you in the post now. $\endgroup$ – Jeff Faraci Apr 13 '14 at 2:36
  • $\begingroup$ @RandomVariable Okay, although I am not sure I see what you're talking about. Thanks though. $\endgroup$ – Jeff Faraci Apr 13 '14 at 2:38
  • $\begingroup$ Nevermind. Your integral involves cosine, not sine. Sorry. $\endgroup$ – Random Variable Apr 13 '14 at 2:45
  • $\begingroup$ @RandomVariable Okay, no problem. Thanks though, as always on your help towards my integrals. The integral I have just posted has a sine in it (similar to this one, but generalized to nth power), maybe you will know that one. Thanks. $\endgroup$ – Jeff Faraci Apr 13 '14 at 2:48
4
$\begingroup$

Using the principal brach of $\log z$, $$\log(1+e^{2ix}) = \log(e^{i x}(e^{-ix}+ e^{i x})) = \log(e^{ix})+ \log(2 \cos x) = ix + \log(2 \cos x) .$$

Squaring both sides,

$$ \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx = \int_{0}^{\pi /6} \Big( ix + \log(2 \cos x) \Big)^2 \ dx .$$

Then equating the real parts on both sides of the equation and rearranging,

$$ \begin{align} \int_{0}^{\pi/3} \log^{2} \left( 2 \cos \frac{x}{2}\right) \ dx &= 2 \int_{0}^{\pi /6} \log^{2}(2 \cos x) \ dx \\ &= 2 \int_{0}^{\pi /6} x^{2} \ dx + 2 \ \text{Re} \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx \\ &= \frac{\pi^{3}}{324} + 2 \ \text{Re} \int_{0}^{\pi /6} \log^{2}(1+e^{2ix}) \ dx . \end{align}$$

Now make the substitution $z = e^{2ix}$.

Then

$$\int_{0}^{\pi/3} \log^{2} \left( 2 \cos \frac{x}{2} \right) \ dx = \frac{\pi^{3}}{324} + \text{Re} \frac{1}{i} \int_{C} \frac{\log^{2}(1+z)}{z} \ dz$$

where $C$ is a portion of the unit circle in the first quadrant of the complex plane.

But since we're using the principal branch of $\log z$, $\log(1+z)$ is analytic on the complex plane for $\text{Re}(z) > -1$.

So the path doesn't matter.

And therefore

$$ \int_{0}^{\pi /3} \log^{2}\left( 2 \cos \frac{x}{2} \right) \ dx = \frac{\pi^{3}}{324} + \text{Re} \frac{1}{i} \int_{0}^{e^{i \pi/3}} \frac{\log^{2}(1+z)}{z} \ dz .$$

You can find an antiderivative of the integrand in terms of polylogarithms by integrating by parts twice.

$$ \begin{align} \int \frac{\log^{2}(1+z)}{z} \ dz &= \log^{2}(1+z)\log(-z) - 2 \int \frac{\log(1+z) \log(-z)}{z} \ dz \\ &= \log^{2}(1+z) \log(-z) + 2 \text{Li}_{2}(1+z) \log(1+z) - 2 \int \frac{\text{Li}_{2}(1+z)}{1+z} \ dz \\ &= \log^{2}(1+z) \log(-z) + 2 \text{Li}_{2}(1+z) \log(1+z) - 2 \text{Li}_{3}(1+z) + C \end{align} $$

Evaluating the integral at the limits and then simplifying a bit,

$$ \int_{0}^{\pi /3} \log^{2}\left( 2 \cos \frac{x}{2} \right) \ dx = \frac{7 \pi^{3}}{324} - \frac{\pi}{6} \log^{2}(3) + \log(3) \text{Im} \ \text{Li}_{2}(1+e^{i \pi /3}) + \frac{\pi}{3} \text{Re} \ \text{Li}_{2}(1+e^{i \pi /3}) $$

$$ - 2 \ \text{Im} \ \text{Li}_{3}(1+e^{i \pi /3}) \approx 0.439089177455491 .$$

$\endgroup$
  • $\begingroup$ I have a question. My integral is from 0 to $\pi/3$. How is yours identical since in your 5th equation, you start with $$ \int_0^{\pi/6} \ln^2( 2\cos x) dx, $$ but my integral is $$ \int_0^{\pi/3} \ln^2( 2\cos x) dx. $$ The integrand doesn't seem to have symmetry from 0 to $\pi/6$, so I don't know how they are equivalent. Thanks. Or, does everything you did apply to my integral as well, you just solved a different integral? $\endgroup$ – Jeff Faraci Apr 13 '14 at 13:29
  • $\begingroup$ It says in the original post that the integral is $\int_{0}^{\pi /3} \ln^{2}(2 \cos \frac{x}{2}) \ dx = 2 \int_{0}^{\pi/6} \ln^{2}(2 \cos x) \ dx$. $\endgroup$ – Random Variable Apr 13 '14 at 13:52
  • $\begingroup$ Agh, I see. I wasn't including the x/2 factor. Great solution. I am going through the details, if I have another question, I may ask you. But for now, thanks again. $\endgroup$ – Jeff Faraci Apr 13 '14 at 13:57
  • $\begingroup$ I'm unfamiliar with the notation $\text{Gl}_{2,1}$. $\endgroup$ – Random Variable Apr 13 '14 at 14:26
  • 1
    $\begingroup$ I evaluated the integral at the limits and simplified a bit. It might not quite be in the form you seek, but numerically it appears to be correct. $\endgroup$ – Random Variable Apr 13 '14 at 15:40

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.