Is the Limit of Riemann Function Product Non-Zero or Finite?

[eljose79]

let be the product R(s)R(s+a) with a a complex or real number..the i would like to know the limit Lim(s tends to e) being e a number so R(e)=0 ¿is there a number a so the limit is non-zero nor infinite?..thanks.

 

[shmoe]

It will only potentially be non-zero when $$\zeta(s)$$ has a pole at e+a. Of course this only happens when e+a=1. This limit will never be infinite, since this lone pole of $$\zeta(s)$$ is simple.

I say "potentially" non-zero, because the zero could have order 2 or more, making the limit zero. All known zeros to date have been simple, and it's suspected that they all are.

 

[eljose79]

Another question..could be proved that 1/R(s+1/2)=O(R(s)?

 

[shmoe]

Another question..could be proved that 1/R(s+1/2)=O(R(s)?

You mean $$\displaystyle\frac{1}{\zeta(1/2+it)}=O(\zeta(1/2+it))$$? If so, no, since the left side has infinitely many poles.

Do you mean something else? I'm not sure what values of s you're considering, and the +1/2 makes me think of the critical line, hence my guess to your meaning.

 

[eljose79]

thanks again... the equality 1/R(1/2+it)=O(R(1/2+it) is true?..

another question let be the Riemann zeta function inside the criticla strip 0<sigma<1 then ..is the product R(a+s)R(s) bounded in the sense exist a and b so a<[R(s+a)R(s)]<b where [x] is the modulus of x [x]=sqrt(x*.x) ?

 

[shmoe]

thanks again... the equality 1/R(1/2+it)=O(R(1/2+it) is true?..

Absolutely not. $$1/\zeta(1/2+it)$$ has poles for infinitely many values of t.

another question let be the Riemann zeta function inside the criticla strip 0<sigma<1 then ..is the product R(a+s)R(s) bounded in the sense exist a and b so a<[R(s+a)R(s)]<b where [x] is the modulus of x [x]=sqrt(x*.x) ?

You're going to have to clarify what you mean by sqrt(x*.x).

There are bounds of the form $$\zeta(\sigma+it)=O(|t|^{k})$$ for any $$\sigma$$ fixed.The exponent k depends on $$\sigma$$. For example, if $$\sigma=1/2$$, then $$k=1/6$$ will work (something slightly smaller than 1/6 is known to be true). If $$\sigma >1$$ then k=0 will work.