Register to reply

A very intriguing integral

by riemannian
Tags: integral, intriguing
Share this thread:
riemannian
#1
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 ??
Phys.Org News Partner Science news on Phys.org
NASA team lays plans to observe new worlds
IHEP in China has ambitions for Higgs factory
Spinach could lead to alternative energy more powerful than Popeye
riemannian
#2
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]


Register to reply

Related Discussions
Intriguing thought experiment Classical Physics 1
Intriguing Question Special & General Relativity 5
Quite Intriguing. What I'm learning. General Discussion 2
Very intriguing stuff... General Discussion 14
Intriguing Statistical/Probability Question Set Theory, Logic, Probability, Statistics 0