A really interesting logarithm integral

I need to prove that

$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x = \frac{2}{9}\pi^2-\frac{1}{3}\psi'(1/3)$$

My approach

We know that

$$\int^1_0 \frac{\log x}{x^2-2\cos(\theta)x+1}\mathrm{d}x =- \frac{\mathrm{cl}_2(\theta)}{\sin(\theta)}$$

Let $\theta=\pi/3$

$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x =- \frac{2}{\sqrt{3}}\mathrm{cl}_2(\pi/3)$$

By definition

\begin{align} \mathrm{cl}_2(\pi/3) &= -\int^{\pi/3}_0 \log(2\sin \frac{t}{2})\mathrm{d}t\\ & = -\frac{\pi}{3}\log(2)-2\pi \int^{1/6}_0\log(\sin(\pi x))\,\mathrm{d}x\\ & = -\frac{\pi}{3}\log(2)-2\pi \int^{1/6}_0\log(\pi)-\log\Gamma(x)-\log\Gamma(1-x)\mathrm{d}x\\ & = -\frac{\pi}{3}\log(2\pi)+2\pi \int^{1/6}_0\log\Gamma(x)+\log\Gamma(1-x)\mathrm{d}x\\ & = -\frac{\pi}{3}\log(2\pi)+2\pi \int^{1/6}_0\log\Gamma(x)+2\pi\int^{1}_{5/6}\log\Gamma(x)\mathrm{d}x \end{align}

Now use the loggamma integral

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z)$$

Implies the two integrals

$$\int_{0}^{1/6} \log \Gamma(x) \, \mathrm dx = \frac{1}{12} \log(2 \pi) + \frac{5}{72} - \zeta^{'}(-1) + \zeta^{'}(-1,1/6)$$

$$\int_{5/6}^{1} \log \Gamma(x) \, \mathrm dx =\frac{1}{2}\log(2\pi)-( \frac{5}{12} \log(2 \pi) + \frac{5}{72} - \zeta^{'}(-1) + \zeta^{'}(-1,5/6))$$

Hence we have

$$\mathrm{cl}_2(\pi/3) = -\frac{\pi}{3}\log(2\pi)+2\pi (\frac{1}{6}\log(2\pi)+\zeta^{'}(-1,1/6)-\zeta^{'}(-1,5/6)) \\=2\pi (\zeta^{'}(-1,1/6)-\zeta^{'}(-1,5/6))$$

This implies that

$$\int^1_0 \frac{\log x}{x^2-x+1}\mathrm{d}x = \frac{-4\pi}{\sqrt{3}}\left(\zeta^{'}(-1,1/6)-\zeta^{'}(-1,5/6) \right) = \frac{2}{9}\pi^2-\frac{1}{3}\psi'(1/3)$$

Maybe using $$\zeta(s,p/q)=2\Gamma(1-s)(2\pi q)^{s-1}\sum_{n=1}^q \sin \left[\frac{\pi s}{2}+ \frac{2 \pi n p}{q}\right] \zeta(1-s,n/q)$$

Question

1. My approach seems to be so indirect I am really interested in seeing more directed approaches that don't use loggamma integral.

2. I am not sure about the last step (relating the derivative of the Hurwitz zeta function to the trigamma.

• Here is a technique handles more general integrals. Commented Aug 30, 2016 at 1:49

Hint. One may write $$\int^1_0 \frac{\log x}{x^2-x+1}dx=\int^1_0 \frac{(1+x)\log x}{1+x^3}dx=\left.\partial_s\int^1_0 \frac{(1+x)\:x^s}{1+x^3}dx\right|_{s=0}$$ then after a change of variable one may use $$\int^1_0 \frac{t^s}{1+t}dt=\frac12\psi\left(\frac{s}2+1\right)-\frac12\psi\left(\frac{s}2+\frac12\right), \quad s>-1.$$