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A very intriguing integral

  1. Mar 1, 2012 #1
    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{1-q^{2}}{1-q^{-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. jcsd
  3. Mar 1, 2012 #2
    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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