- #1
JamesZhu
- 1
- 0
- TL;DR Summary
- make a bridge between complex function and real function
For original Zeta function, ζ(s)=1+1/2^s+1/3^s+1/4^s+... =1+e^(-slog2)+e^(-slog3)+e^(-slog4)+... , Re(s)>1
Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers.
So we only handle with original Zeta function. All Zeta functions mentioned later are in original definition and Re(s)>1.
If we regards ζ(s) as some function got from Laplace transformation, and let this real function be ζ(x), that means L[ζ(x)]=ζ(s), then:
ζ(x)=L^-1[ζ(s)]=δ(x)+δ(x-log2)+δ(x-log3)+δ(x-log4)+... , this represents a series of Dirac delta functions at the points of x=0, log2, log3, log4, ... ,
It may be still difficult to understand what ζ(x) means, but once it is integrated, the truth is clear:
ζ1(x)=∫ζ(x)dx=u(x)+u(x-log2)+u(x-log3)+u(x-log4)+,,, ,
This represents a combination of unit step functions start at the points x=0, ln2, ln3, ln4, ... , obviously has an exact upper bound function f2(x)=e^x and a lower bound function f1(x)=e^x-1. And it is easily to find L[ζ1(x)]=ζ(s)/s, L[e^x]=1/(s-1), L[e^x-1]=1/(s-1)-1/s=s/(s-1).
This is why many of the series decomposition expressions of ζ(s) contain 1/(s-1), because the main item of ζ1(x) and ζ(x) both are e^x.
It make a clear picture of Zeta function which haven`t got before by other ones, and will lead to more interesting and important results.
Riemann extended the Zeta function to the region where s≠1 using analytical extension. New Zeta function is in the form of contour integration, which appears simple but is actually more inconvenient to analyze than the original Zeta function. The original Zeta function already contains all the information about the distribution of prime numbers.
So we only handle with original Zeta function. All Zeta functions mentioned later are in original definition and Re(s)>1.
If we regards ζ(s) as some function got from Laplace transformation, and let this real function be ζ(x), that means L[ζ(x)]=ζ(s), then:
ζ(x)=L^-1[ζ(s)]=δ(x)+δ(x-log2)+δ(x-log3)+δ(x-log4)+... , this represents a series of Dirac delta functions at the points of x=0, log2, log3, log4, ... ,
It may be still difficult to understand what ζ(x) means, but once it is integrated, the truth is clear:
ζ1(x)=∫ζ(x)dx=u(x)+u(x-log2)+u(x-log3)+u(x-log4)+,,, ,
This represents a combination of unit step functions start at the points x=0, ln2, ln3, ln4, ... , obviously has an exact upper bound function f2(x)=e^x and a lower bound function f1(x)=e^x-1. And it is easily to find L[ζ1(x)]=ζ(s)/s, L[e^x]=1/(s-1), L[e^x-1]=1/(s-1)-1/s=s/(s-1).
This is why many of the series decomposition expressions of ζ(s) contain 1/(s-1), because the main item of ζ1(x) and ζ(x) both are e^x.
It make a clear picture of Zeta function which haven`t got before by other ones, and will lead to more interesting and important results.
