i've been trying to do this integral , but with no luck .
$\int_1^\infty \frac{\left \{x \right \}}{x}\left(\frac{1}{x^{s}-1}\right) \; dx$
$\Re(s)>1 $ , $\left \{x \right \} $ is the fractional , sawtooth function.
i have tried the Fourier expansion of the sawtooth function :
$ \int_{1}^\infty \frac{\left \{x \right \}}{x}\left(\frac{1}{x^s - 1}\right)\; dx = \int_1^\infty \frac{1}{x(x^{s}-1)}\left(\frac{1}{2}-\frac{1}{\pi}\sum_{n=1}^\infty \frac{\sin(2\pi i nx)}{n} \right) \; dx$
$=\int_1^\infty\frac{1}{x(x^s-1)}\left(\frac{1}{2}+\frac{1}{2\pi i}\ln \left(\frac{1-q^2}{1-q^{-2}} \right)\right ) \; dx$
where $ q $ is the nome : $ q=e^{i \pi x}$
after some manipulation , the integral reduces to :
$\frac{1}{2\pi i }\int_1^\infty \frac{\left(\pi i +\ln(-e^{2\pi i x}) \right )}{x(x^s - 1)} \; dx$
but that brought me no where near a solution !! any suggestions on how to do the integral ??