Let for $\Re s>0$ $$\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^{s}}$$ the Dirichlet eta function. After I read Example 5 and Corollary 9 from [1], I've asked myself an analogue

Question. If I define for $t\in \left( 0,\infty \right) $ (our $x$ inside the integral is thus a positive real) $$f(t):=-\int_0^t\Re\frac{\eta' \left( \frac{1}{2}+ix \right) }{\eta \left( \frac{1}{2}+ix \right) }dx,$$ a) is it well-defined?, and b) what's about the convexity of this function for $t\in \left( 0,\infty \right) $?

Since I know the calculations of professors in [1] to get such the analogue in which I was inspired, then as an answer for b) if you want only is required the key hint to study/deduce what's about the convexity ($f''(t)\geq 0$, $\forall t>0$) of our $f(t)$. Many thanks.

Was fixed a typo concerning the condition to be convex.


[1] Arias de Reyna and Van de Lune, On the exact location of the non-trivial zeros of Riemann's Zeta function, Acta Arithmetica 163.3 (2014).

  • 1
    $\begingroup$ @SimpleArt Why do you believe that $n^{1/2+ix}=e^{(1/2+ix)\log(n)}$ is not well-defined? We assume the principal branch for the logarithm here. $\endgroup$ – Mark Viola Nov 27 '16 at 15:55
  • 1
    $\begingroup$ You may wish to see that $\eta(s)=(1-2^{1-s})\zeta(s)$, where $\zeta(s)$ is the Riemann zeta function, and more information can be found concerning the Wiki for the Riemann hypothesis. $\endgroup$ – Simply Beautiful Art Nov 27 '16 at 16:03
  • 1
    $\begingroup$ Notice that $$\int\Re\frac{\eta'(1/2+ix)}{\eta(1/2+ix)}dx=\Re\int\frac{\eta'(1/2+ix)}{\eta(1/2+ix)}dx=\Re i\log\left[\eta(1/2+ix)\right]=-\arg\left[\eta(1/2+ix)\right]$$ $\endgroup$ – Simply Beautiful Art Nov 27 '16 at 16:14
  • 1
    $\begingroup$ While not an answer, the following puts some good intuition that this is true: wolframalpha.com/input/?i=arg(dirichlet+eta(1%2F2%2Bix)) $\endgroup$ – Simply Beautiful Art Nov 27 '16 at 16:19
  • 1
    $\begingroup$ You have to be careful here. $\eta'/\eta$ has many poles on $Re(s) = 1/2$, so $\int_0^t \frac{\eta'}{\eta}(1/2+ix)dx$ can be defined only as a principal value. And in that case yes $\int_0^t \frac{\eta'}{\eta}(1/2+ix)dx = -\text{arg} (\eta(1/2+ix))$ with the branch of $\text{arg}[\eta(s)]$ continuous (and monotone) on $Re(s) = 1/2$ and $\text{arg}( \eta(1/2)) =0$. Anyway this is mainly a tautology. @SimpleArt $\endgroup$ – reuns Nov 29 '16 at 5:06

Your Answer

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