# Evaluate $\int_{0}^{\frac {\pi}{3}}x\log(2\sin\frac {x}{2})\,dx$

Prove that $$\int_0^{\pi/3}x\log \left(2 \sin\frac {x}{2}\right)\,dx = \frac {2\zeta(3)}{3}-\frac {\pi^2}{9}\log (2\pi)+\frac {2\pi ^2}{3}\log \left|\frac {\Gamma_2 \left(\frac {5}{6}\right)}{\Gamma_2 \left(\frac {7}{6}\right)}\right|$$

(Editor's note, July 2019: According to Adamchik's paper "On the Barnes Function" (p. 4), the notation $$\Gamma_n(z)$$ follows Vigneras and Vardi and is the reciprocal of the multiple gamma function $$G_n$$,

$$\Gamma_n(z) = \frac{1}{G_n(z)}$$

with the special case,

$$\Gamma_2(z) = \frac{1}{G_2(z)} = \frac{1}{G(z)}$$

where $$G(z) = G_2(z)$$ is the double Gamma function aka Barnes G function.

• Alternatively, $$3\int_0^{\pi/3}x\ln\left(2\sin\frac{x}2\right)\,dx=-\pi\,\color{red}\kappa+2\,\zeta(3)\\ 2\int_0^{\pi/2} x\ln\left(2\sin \frac{x}2\right)\,dx =-\pi\,K+\frac{35}{16}\zeta(3)$$ where Gieseking's constant $\color{red}\kappa =\rm{Cl}_2\big(\tfrac\pi3\big)$ is just the cubic counterpart of Catalan's constant $K = \rm{Cl}_2\big(\tfrac\pi2\big)$. Jul 1, 2019 at 17:18

I don't know what a double Gamma function is, so I will use the Barnes G function. Use the fourier expansion of $\log\sin x$ which is $$\log\sin x=-\log2-\sum_{k=1}^\infty\frac{\cos 2kx}{k}$$ Then the rest is pretty straightforward. \begin{align} \int_0^{\pi/3}x\log\left(2\sin\frac x2\right)\,dx&=-\int_0^{\pi/3}x\sum_{k=1}^\infty\frac{\cos kx}{k}\,dx\\&=-\sum_{k=1}^\infty\int_0^{\pi/3}\frac xk\cos kx\,dx\\&=-\sum_{k=1}^\infty\left(\frac{\pi}{3k^2}\sin\frac{k\pi}{3}+\frac{1}{k^3}\cos\frac{k\pi}{3}-\frac{1}{k^3}\right)\\&=-\frac{\pi}{3}\operatorname{Cl}_2\left(\frac\pi3\right)-\operatorname{Cl}_3\left(\frac\pi3\right)+\zeta(3) \end{align} where $\operatorname{Cl}_n(x)$ are the Clausen function. Let $a_n$ be $(-1)^{n/3}$ if $n$ is a multiple of $3$ and otherwise zero. Then we have \begin{align}\operatorname{Cl}_3\left(\frac\pi3\right)&=\sum_{n=1}^\infty\frac{1}{n^3}\cos\frac{n\pi}{3}=\sum_{n=1}^\infty\frac{1}{n^3}\left(\frac{(-1)^{n-1}}{2}+\frac{3}{2}a_n\right)\\&=\frac12\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^3}-\frac{3}{2}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(3n)^3}\\&=\frac{\zeta(3)}{3}\end{align} and from the relation of the clausen function and the Barnes G function, we have $$\operatorname{Cl}_2\left(\frac\pi3\right)=2\pi\log\left(\frac{G(5/6)}{G(7/6)}\right)+\frac\pi3\log(2\pi)$$

Plugging altogether gives $$\int_0^{\pi/3}x\log\left(2\sin\frac x2\right)\,dx=\frac{2\zeta(3)}{3}-\frac{\pi^2}9\log(2\pi)-\frac{2\pi^2}{3}\log\left(\frac{G(5/6)}{G(7/6)}\right)$$

• Upon further research, it turns out the notation $\Gamma_n(z)$ is the reciprocal of the multiple gamma function $G_n(z)$ with the special case $G(z) = G_2(z)$ being your Barnes G function. Hence, what I thought was a sign error in the OP's $\log(n)$ was just a reciprocal. I've added a link to Adamchik's paper in the OP's post as reference. Jul 25, 2019 at 3:42

$\def\Li{{\rm Li}}$Not an answer yet, but maybe it could help (I hope...).

From Fourier analysis we have $$\ln\left(2\sin \frac{x}{2}\right)=-\sum_{n=1}^{+\infty}\frac{\cos(nx)}{n}$$ Therefore \begin{align} \int_0^{\pi/3}x\ln\left(2\sin \frac{x}{2}\right)\,dx&=-\int_0^{\pi/3}x\sum_{n=1}^{+\infty}\frac{e^{inx}+e^{-inx}}{2n}\,dx\\ &=-\sum_{n=1}^{+\infty}\int_0^{\pi/3}\frac{xe^{inx}+xe^{-inx}}{2n}\,dx\\ &=-\frac{1}{2}\sum_{n=1}^{+\infty}\left(\left.\frac{xe^{inx}}{in^2}\right|_{x=0}^{\pi/3}-\int_0^{\pi/3}\frac{e^{inx}}{in^2}\,dx-\left.\frac{xe^{-inx}}{in^2}\right|_{x=0}^{\pi/3}+\int_0^{\pi/3}\frac{e^{-inx}}{in^2}\,dx\right)\\ &=-\frac{1}{2}\left(\frac{\pi}{3i}\Li_2\left(e^{\frac{\pi i}{3}}\right)+\Li_3\left(e^{\frac{\pi i}{3}}\right)-\Li_3(1)-\frac{\pi}{3i}\Li_2\left(e^{-\frac{\pi i}{3}}\right)+\Li_3\left(e^{-\frac{\pi i}{3}}\right)-\Li_3(1)\right)\\ &=-\frac{1}{2}\left(\frac{\pi}{3i}\left[\Li_2\left(e^{\frac{\pi i}{3}}\right)-\Li_2\left(e^{-\frac{\pi i}{3}}\right)\right]+\Li_3\left(e^{\frac{\pi i}{3}}\right)+\Li_3\left(e^{-\frac{\pi i}{3}}\right)-2\zeta(3)\right) \end{align} Interestingly, from Wolfram Alpha I get the following results \begin{align} \Li_2\left(e^{\frac{\pi i}{3}}\right)-\Li_2\left(e^{-\frac{\pi i}{3}}\right)&=\frac{i\sqrt{3}}{36}\left[\psi^{(1)}\left(\frac{1}{6}\right)-\psi^{(1)}\left(\frac{5}{6}\right)+\psi^{(1)}\left(\frac{1}{3}\right)-\psi^{(1)}\left(\frac{2}{3}\right)\right]\\[12pt] \Li_3\left(e^{\frac{\pi i}{3}}\right)+\Li_3\left(e^{-\frac{\pi i}{3}}\right)&=\frac{2\zeta(3)}{3} \end{align} Substituting the above results, we get \begin{align} \int_0^{\pi/3}x\ln\left(2\sin \frac{x}{2}\right)\,dx &=\frac{2\zeta(3)}{3}+\frac{\pi}{72\sqrt{3}}\left[\psi^{(1)}\left(\frac{5}{6}\right)-\psi^{(1)}\left(\frac{1}{6}\right)+\psi^{(1)}\left(\frac{2}{3}\right)-\psi^{(1)}\left(\frac{1}{3}\right)\right] \end{align}

• I noticed that $$\psi^{(1)}\left(\frac56\right)-\psi^{(1)}\left(\frac16\right)+\psi^{(1)}\left(\frac23\right)-\psi^{(1)}\left(\frac13\right)= \\6\left(\psi^{(1)}\left(\frac23\right)-\psi^{(1)}\left(\frac13\right)\right)$$ After that using $(7)$ and $(5)$ from here you get this is $-36\sqrt3\operatorname{Cl}_2\left(\frac{2\pi}{3}\right)$. Using this connection between Barnes G-function and Clausen function you will get the solution. Oct 1, 2014 at 12:57
• @user153012 What kind of identity did you use to get $$\psi^{(1)}\left(\frac56\right)-\psi^{(1)}\left(\frac16\right)+\psi^{(1)}\left(‌​\frac23\right)-\psi^{(1)}\left(\frac13\right)= \\6\left(\psi^{(1)}\left(\frac23\right)-\psi^{(1)}\left(\frac13\right)\right)\, ?$$ Oct 1, 2014 at 13:00
• I just noticed, or observed it. I don't know how to prove it. Maybe it comes somehow from the multiplication theorem of polygamma function, or I don't know. Maybe I will take a question on it. Oct 1, 2014 at 13:08
• @user153012 You are right. It does follow from the multiplication theorem. Oct 1, 2014 at 15:21