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


  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.


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 $$ \int^1_0 \frac{t^s}{1+t}dt=\frac12\psi\left(\frac{s}2+1\right)-\frac12\psi\left(\frac{s}2+\frac12\right), \quad s>-1. $$

  • $\begingroup$ Sometimes I mess the easiest tricks. $\endgroup$ – Zaid Alyafeai Aug 29 '16 at 20:45
  • 1
    $\begingroup$ Nice and tricky (+1) $\endgroup$ – Behrouz Maleki Aug 29 '16 at 20:52

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.