Sharpening a bound on $\zeta'(s)$ 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.
 A: 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.
