# LogSine Integral $I=-\int_0^{\pi/3} \ln^2\big(2\cos \frac{\theta}{2}\big) d\theta$

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

• What's the closed form? – Antonio Vargas Apr 12 '14 at 23:58
• @AntonioVargas i have added the closed form for you in the post now. – Jeff Faraci Apr 13 '14 at 2:36
• @RandomVariable Okay, although I am not sure I see what you're talking about. Thanks though. – Jeff Faraci Apr 13 '14 at 2:38
• Nevermind. Your integral involves cosine, not sine. Sorry. – Random Variable Apr 13 '14 at 2:45
• @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. – Jeff Faraci Apr 13 '14 at 2:48

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,

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

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

• 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? – Jeff Faraci Apr 13 '14 at 13:29
• It says in the original post that the integral is $\int_{0}^{\pi /3} \ln^{2}(2 \cos \frac{x}{2}) \ dx = 2 \int_{0}^{\pi/6} \ln^{2}(2 \cos x) \ dx$. – Random Variable Apr 13 '14 at 13:52
• 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. – Jeff Faraci Apr 13 '14 at 13:57
• I'm unfamiliar with the notation $\text{Gl}_{2,1}$. – Random Variable Apr 13 '14 at 14:26
• 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. – Random Variable Apr 13 '14 at 15:40