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A very intriguing integral

by riemannian
Tags: integral, intriguing
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Mar1-12, 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{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 ??
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Mar1-12, 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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