# Evaluation of :$\sum_{n=1}^{\infty}\frac{\sin n \log n}{n}$

I have tried to evaluate this sum $\sum_{n=1}^{\infty}\frac{\sin n \log n}{n}$ using :$\sum_{n=1}^{\infty}\frac{\sin n }{n}$ , Really the partial sum of the titled series given by polylogarithm function and lerch transcendent functions as shown here, But i'm not familiar about convergence of these function for large n , Then my question here is :

What is the exact value of :$$\sum_{n=1}^{\infty}\frac{\sin n \log n}{n}$$ ?

• Honestly I would not expect a closed form to exist. – Szeto Jul 6 '18 at 14:23
• @gammatester But Wolfy says it is around 0.007. – Szeto Jul 6 '18 at 14:37
• @szeto: Yes, it seems to be non-zero, and therefore I deleted my comment. I have plotted with Maple because here WA does give a result. Plotting up to 2000 shows that it is slightly above 0. – gammatester Jul 6 '18 at 14:44

$$0 : $$\enspace\displaystyle \sum\limits_{n=1}^\infty \frac{\ln n}{n}\sin(2\pi n x) = \frac{\pi}{2}\left(\ln\frac{\Gamma(x)}{\Gamma(1-x)}-(1-2x)(\gamma + \ln(2\pi))\right)$$

see e.g. E. E. Kummer, page 1 to 4

See also here , look for the first derivation for $$t$$ of $$B_t(x)$$ , a generalization of the Bernoulli polynomial and it’s Fourier series.

It follows:

$$\sum\limits_{n=1}^\infty \frac{\sin n\ln n}{n} = \frac{\pi}{2}\left(\ln\frac{\Gamma\left(\frac{1}{2\pi}\right)}{\Gamma\left(1-\frac{1}{2\pi}\right)}-\left(1-\frac{1}{\pi}\right)(\gamma + \ln(2\pi))\right)$$

• nice closed form – zeraoulia rafik Jul 6 '18 at 14:56
• @zeraouliarafik : It has to do with the Fourier series for the logarithm of the Gamma function. :) – user90369 Jul 6 '18 at 15:03
• You mean something like this: en.wikipedia.org/wiki/Gamma_function#Fourier_series_expansion – gammatester Jul 6 '18 at 15:09
• This is amazing. – Szeto Jul 6 '18 at 15:13
• @gammatester : Of course, but I haven't checked Wikipedia or others. It's good that you've mentioned it. – user90369 Jul 6 '18 at 15:13