I want to find an upper bound for $\zeta'(s)$ along a vertical line $\Re(s)=b$, where $-1<b<0$.
One way to do this is using $$\frac{\zeta'(b+iT)}{\zeta(b+iT)}=O_b(\log T)$$ and $$\zeta(b+iT)=O_{b,\varepsilon}(T^{1/2-b+\varepsilon})$$ for each $\varepsilon>0$.
Multiplying them gives us $$\zeta'(b+iT)=O_{b,\varepsilon}(T^{1/2-b+\varepsilon}\log T)$$ as $T\rightarrow\infty$. I want to know if there is a bound, for fixed $b$, that is sharper.
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2$\begingroup$ The functional equation expresses $\zeta'(b+iT)$ in terms of $\zeta(1-b-iT)$, $\zeta'(1-b-iT)$, and the complex Gamma function and its derivative on the lines of real part $b$ and $1-b$. If $-1<b<0$ then $1 < 1-b < 2$ so $\zeta(1-b-iT)$ and $\zeta'(1-b-iT)$ are bounded and it's just a matter of the estimating the relevant Gamma and $\Gamma'$ factors. $\endgroup$– Noam D. ElkiesCommented Mar 10, 2014 at 3:34
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7$\begingroup$ Use the functional equation connecting $\zeta(b+iT)$ to $\zeta(1-b-iT)$ and differentiate. Then use Stirling's formula. This gives the bound $|\zeta^{\prime}(b+it)|= O_b(T^{1/2-b}\log T)$ (your expression seems to have a typo). This is also best possible since $|\zeta(1-b-iT)|$ is bounded away from zero. $\endgroup$– LuciaCommented Mar 10, 2014 at 3:34
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$\begingroup$ Yes, you are correct about the typo, so I fixed that. Thanks for letting me know that this is the best possible estimate. $\endgroup$– B.W.Commented Mar 10, 2014 at 3:40
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$\begingroup$ Here you can find some bounds math.univ-lille1.fr/~ramare/TME-EMT/Articles/Art06.html $\endgroup$– user21574Commented Mar 10, 2014 at 13:27
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1 Answer
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I'm not sure if you need me to go further, but if I were you I'd start out with the functional equation. Take the derivative of both sides of the functional equation and you can derive:
$\frac{\zeta'(1-s)}{\zeta(1-s)} = log(2\pi)+\frac{\pi}{2}\tan(\frac{\pi s}{2}) - \psi(s)-\frac{\zeta'(s)}{\zeta(s)}$.
Letting $s = 1-b-iT$, the left hand side will be exactly what you're looking at.
I am some what confused. You say the vertical line where $Re(s)=b$ where $0<b<1$. If it's a vertical line, then wouldn't $T$ vary and $b$ be held fixed?
If $T$ is allowed to vary (i.e. a vertical line), then wouldn't the answer depend on whether the line crosses the real axis? I believe $\tan(\frac{\pi}{2}s)$ blows up in magnitude when $b$ is really small in absolute value.
If $T$ is held fixed, and you allowed $b$ to vary, then believe the $\tan$ term is bounded since you'd be a fixed distance away from the problem point $b=0$. If this is the case, Lucia is correct I believe.
Perhaps I'm wrong...it is awful late and my mind is worn out from all day of flying across the country. I apologize if I made a mistake somewhere in my logic.
If I am wrong in my logic, could someone explain? I'm very interested as well.
