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

Question

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 :$$\sum_{n=1}^{\infty}\frac{\sin n \log n}{n}$$ ?

Answer 1

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

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:

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