
#1
Mar112, 11:29 AM

P: 5

greetings . the following integral appears in some references on analytic number theory . i am really intrigued by it . and would love to understand it .
[tex] \int_{1}^{\infty}\frac{\left \{x \right \}}{x}\left(\frac{1}{x^{s}1}\right)dx[/tex] [itex]\Re(s)>1 [/itex] , [itex]\left \{x \right \} [/itex] is the fractional , sawtooth function . i have tried the Fourier expansion of the sawtooth function : [tex] \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{1q^{2}}{1q^{2}} \right)\right )dx[/tex] where [itex] q [/itex] is the nome : [tex] q=e^{i \pi x}[/tex] but that brought me no where near a solution !! any suggestions on how to do the integral ?? 



#2
Mar112, 01:54 PM

P: 5

after some manipulation , the integral reduces to :
[tex] \frac{1}{2\pi i }\int_{1}^{\infty}\frac{\left(\pi i +\ln(e^{2\pi i x}) \right )}{x(x^{s}1)}dx[/tex] 


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