- #1
riemannian
- 5
- 0
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 ??
\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 ??
