4
$\begingroup$

Let the complex variable $s=\sigma+it$, then from the following identity valid for $\sigma=\Re s>1$ $$\zeta(s)=s\int_1^\infty \frac{[x]}{x^{s+1}}dx$$ where $\zeta(s)$ is the Riemann Zeta function, I've computed by differentition under the integral sign $$ \frac{\zeta(s)}{s}- \frac{d\zeta(s)}{d\sigma} =s(\sigma+1)\int_1^\infty\frac{[x]\log x}{x^{\sigma+2}}e^{-it\log x} dx.$$

I don't know if my deduction was right, and I believe that previous makes sense for $\sigma>2$, and on the strip $ \left\{ x+iy:a\leq x\leq b \right\} $ thus with $a>2$, since $\frac{\zeta(s)}{s}- \frac{d\zeta(s)}{d\sigma}$ is a bounded function of $s$ defined on the strip, holomorphic in the interior of the strip and continuous on the whole strip, defining $$M(x)=\sup_y \left| \frac{\zeta(x+iy)}{x+iy}- \frac{d\zeta(x+iy)}{dx} \right| ,$$ the Hadamard three-lines theorem proves that $\log M(x)$ is a convex function on $\left[ a,b \right]$ (here as I've said $a>2$).

Question. Were rights, do make sense all my claims looking a simple application of the Hadamard three-lines theorem for a function on a strip to the right of $\sigma>2$? Thanks in advance.

$\endgroup$
  • $\begingroup$ If there are no mistakes, I know by comments from an user in this site, that derivation under the integral sign should be justified since everything converges absolutely, writing the limit defining the derivative explicity, but I would like see details to learn these appreciated computations. $\endgroup$ – user243301 May 19 '16 at 18:11
  • 1
    $\begingroup$ there is a big problem with your "derivative" : $\frac{d\zeta(s)}{ds} = \int_1^\infty \lfloor x \rfloor x^{-s-1} dx + \int_1^\infty \lfloor x \rfloor x^{-s-1} (\ln x) dx$. did you understand the derivatives of holomorphic/analytic functions, the Cauchy-Riemann equations ? and $a > 1$ is enough for applying Hadamard's 3 lines theorem $\endgroup$ – reuns May 21 '16 at 10:03
  • $\begingroup$ I think I've understood the basics of the theory of derivatives of analytic functions , but not from the point of view to put in practice satisfactorily, so I try to get my hands simple examples to refresh and learn what are my mistakes. I don't know why $a>1$ is suffcient. Then I am agree that my computations are wrong or aren't the best. Thanks @user1952009 $\endgroup$ – user243301 May 21 '16 at 11:15
  • 1
    $\begingroup$ there is a typo, I forgot a $s$ term : $\zeta'(s) = (s\int_1^\infty \lfloor x \rfloor x^{-s-1} dx)' = \int_1^\infty \lfloor x \rfloor x^{-s-1} dx + s\int_1^\infty \lfloor x \rfloor x^{-s-1} (\ln x) dx$ $\endgroup$ – reuns May 24 '16 at 13:52
  • 1
    $\begingroup$ the right formula is $\frac{d}{ds} x^{-s-1} = - x^{s-1} \log x$ for every $s \in \mathbb{C}$ and $x \ne 0$, because $s \to x^{-s-1} = e^{- (s+1) \log x}$ is analytic in $s$, hence we don't need partial derivatives $\frac{\partial}{\partial Re(s)}x^{-s-1}$ and $\frac{\partial}{\partial Im(s)}x^{-s-1}$ because the two are related by a constant factor $i$ (the Cauchy-Riemann equations) $\endgroup$ – reuns May 24 '16 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy