Bounding a number-theoretic integral

Question

Find a good upper bound on $$\int_1^T\frac{\zeta'(s)}{\zeta(s)\zeta(1-s)}X^sdt,$$ where $s=c+it$ for a constant $c>1$ and $X>0$ is a parameter. If needed, we can assume RH.

My attempt here is to use the functional equation of $\zeta$, namely $\zeta(s)=\chi(s)\zeta(1-s)$ for a specific function $\chi$. Using this, we see $\chi(s)=\frac{\zeta(s)}{\zeta(1-s)}$. This converts our integral to $$\int_1^{T}\frac{\zeta'(s)}{\zeta(s)}\chi(s)X^sdt.$$ Further, we know asymptotically (due to Stirling) that for large $t$, $$\chi(c+it)\sim (t/2\pi)^{1/2-c-it}e^{it+i\pi/4},$$ so my next step was to use the bound $$\left|\frac{\zeta'}{\zeta}(c+iT)\right|=O(\log^2T)$$ together with $|X^s|=X^c$ and the asymptotic for $\chi$. However this fails; the asymptotic works for large $t$ yet our integral runs from $1$ up to $T$, so this approach doesn't work. Is there another way to do this?

Answer 1

The integral looks something like $$\sum _{n=1}^\infty \frac {\Lambda (n)}{n^c}\int _1^Tt^{1/2-c}\cdot e(t-t\log (X/nt))\cdot dt\hspace {10mm}e(z)=e^{2\pi iz}.$$

The derivative of the phase is something like $\log t\gg 1$ and oscillatory integrals are bounded by the derivative (see earlier lemmas of Ch4 of Titchmarsh), so here we'd like to say that the integral is $$\ll T^{1/2-c}$$ which is a big saving on the absolute value bound $\ll T^{3/2}$. The problem is that I spoke too vaguely about the derivative - looking at it more carefully we see that it is $\log (X/nt)$ so there's a problem if this gets close to zero. In that case you have to analyse the integral more precisely around $t=X/n$, a process called the method of stationary phase (see slightly later lemmas of Ch4 of Titchmarsh). You should get something like the square-root of the second derivative, so here $\sqrt t$, giving you a total bound for the integral $\ll T^{1-c}$ which is still $\sqrt T$ better than the absolute value bound.