Register to reply

Trivial zeros in the Riemann Zeta function

by msariols
Tags: riemann, riemann zeta, zeta function
Share this thread:
Oct8-11, 11:46 AM
P: 1
Hello, I have read in many articles that the trivial zeros of the Riemann zeta function are only the negative even integers (-2, -4, -6, -8, -10, ...).

The reason why these are the only ones is that when substituting them in the functional equation, the function is 0 because sin([itex]\frac{x\pi}{2}[/itex])=0.

My question is: why aren't positive even integers trivial zeros too?

The sinus of k[itex]\pi[/itex] =0 with either k[itex]\in[/itex]Z positive or negative.

Remember that the functional equation is:

[itex]\zeta[/itex](x)=[itex]\zeta[/itex](1-x)[itex]\Gamma[/itex] (1-x)2[itex]^{x}[/itex][itex]\pi[/itex][itex]^{x-1}[/itex]sin ([itex]\frac{x\pi}{2}[/itex])
Phys.Org News Partner Science news on
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
Oct8-11, 01:06 PM
P: 1,666
At the even integers, the simple poles of [itex]\Gamma(1-z)[/itex] are canceled by the simple zeros of [itex]\sin(\pi z/2)[/itex] and since the poles and zeros are of the same order (simple), this cancelation is non-zero, that is, the singularity is a removable one. For example consider the limit:

[tex]\lim_{x\to 4} \; \Gamma(1-x) \sin(\pi x/2)=\frac{\pi}{12}[/tex]
Oct25-11, 04:34 PM
P: 150
also because at the positive even integers, the zeta function is defined the Dirichlet series 1+1/2^s+1/3^s+1/4^s+... which converges for all positive even numbers.

Register to reply

Related Discussions
Can someone explain zeros and zeta function for Riemann Hypothesis? (Yr13) Linear & Abstract Algebra 19
Trivial zeros of Zeta Riemann Function Linear & Abstract Algebra 1
Trivial zeros of the Riemann zeta function Linear & Abstract Algebra 3
Riemann Zeta function zeros Calculus 1
Riemann Zeta zeros Linear & Abstract Algebra 31