I ran across a double sum and was wondering if someone may be adept at evaluating it. I must admit that my double summation skills could be better, and I am always ready to learn more.

Show that:


I found this while playing around with:

$$\int_{0}^{1}\frac{\log(1-xe^{\frac{\pi i}{3}})\log(1-xe^{\frac{-\pi i}{3}})}{x}dx$$

I would enjoy seeing a clever evaluation of the sum.

I tried breaking it up as:

$$\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\cos(\frac{\pi k}{3})\cos(\frac{\pi n}{3})}{kn^{2}}$$ $$-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\cos(\frac{\pi k}{3})\cos(\frac{\pi n}{3})}{n^{2}(k+n)}$$ $$+\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\sin(\frac{\pi k}{3})\sin(\frac{\pi n}{3})}{kn^{2}}$$ $$-\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{\sin(\frac{\pi k}{3})\sin(\frac{\pi n}{3})}{n^{2}(k+n)}$$

The third from the top evaluates to $$\frac{\pi}{3}Cl_{2}(\frac{\pi}{3})=\frac{\pi}{3}\left(\frac{\sqrt{3}}{6}\psi_{1}(1/3)-\frac{\pi^{2}\sqrt{3}}{9}\right)$$

I think the top one evaluates to 0.

The other 2 are a little more challenging.

Would anyone enjoy lending a hand and showing a method to evaluate said double sum...or even the integral for that matter?.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.