# A very intriguing integral

1. Mar 1, 2012

### riemannian

greetings . the following integral appears in some references on analytic number theory . i am really intrigued by it . and would love to understand it .
$$\int_{1}^{\infty}\frac{\left \{x \right \}}{x}\left(\frac{1}{x^{s}-1}\right)dx$$

$\Re(s)>1$ , $\left \{x \right \}$ 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 :
$$q=e^{i \pi x}$$

but that brought me no where near a solution !! any suggestions on how to do the integral ??

2. Mar 1, 2012

### riemannian

after some manipulation , the integral reduces to :

$$\frac{1}{2\pi i }\int_{1}^{\infty}\frac{\left(\pi i +\ln(-e^{2\pi i x}) \right )}{x(x^{s}-1)}dx$$