Does the Fourier Transform of Ln\zeta(2e^{-s}) Exist?

[eljose]

Let be the function $$Ln\zeta(2e^{-s})$$ does its Fourier transform exist?...where $$\zeta(s)$$ is teh Zeta function of Riemann...

 

[matt grime]

In LaTeX the command is \log, ans I presume our latex engine supports this: $$\log$$; this this is maths, all logs in mathematics are to base e with the exception of people doing entropy/computing/information theory who use base 2, but would still use log since the usage makes clear that they are in base 2 (no that isn't supposed to start a debate or a flame war, it is a simple observation of the standard notation in higher level maths i.e. NOT what you were told in calc 101).

 

[tacman]

The answer to this question depends on the specific definition of the Fourier transform being used. In general, the Fourier transform is defined for functions that are integrable over the entire real line. However, the function Ln\zeta(2e^{-s}) is not integrable over the entire real line, as it has a singularity at s=0. Therefore, if the Fourier transform is defined in the traditional sense, then the Fourier transform of Ln\zeta(2e^{-s}) does not exist.

However, there are alternative definitions of the Fourier transform that can be used for functions that are not integrable over the entire real line. For example, the tempered distributions approach allows for the Fourier transform of functions that have certain types of singularities. In this case, the Fourier transform of Ln\zeta(2e^{-s}) may exist.

In addition, the Riemann zeta function itself has a complex variable, s, and the function Ln\zeta(2e^{-s}) can be extended to a meromorphic function in the complex plane. In this case, the Fourier transform may exist as a complex function.

Overall, the existence of the Fourier transform of Ln\zeta(2e^{-s}) depends on the specific definition being used and the properties of the Riemann zeta function. Further analysis and clarification of the definitions involved would be necessary to determine the exact answer.