By Stirling we have $\chi(c+it)\sim \left(\frac{t}{2\pi}\right)^{\frac{1}{2}-c-it}e^{it+i\frac{\pi}{4}}$ for large $t$, do you have a reference for this being true for all $t$?

ah yes this is where I was getting confused -- I'm not sure why $\chi(c+it)\approx t^{1/2-c}$ (I assume you mean absolute value?) but in any case, I think the absolute value of chi is like that for large t but in the regime where t is small I'm not sure why that is true - modulus of chi decays quickly so the small t seems to be where our problem lies

Actually how did you deduce in the first place that $\int_{1}^T{X/n}^{it}\chi(c+it)dt=\sum_{n=1}^{\infty}n^{-c}\Lambda(n)\int_1^Tt^{1/2-c}e^{i(t-t\log(X/nt))}$? I agree that we should get a von mangoldt type sum, but I'm not sure where the integral comes from. We know $(X/n)^{it}=e^{it\log(X/n)}$ so our integrand should be $e^{it\log(X/n)}\chi(c+it)$ (or maybe the abs value), but the derivative of the phase is zero so stationary phase doesnt apply (at least if my computation is correct)-- could you elaborate on how you got the integral?

but if we claim that the integral itself is $\ll T^{1/2-c}$ then if $c=1.2$ for example we are saying the integral is $\ll T^{-0.7}$ which decays as $T$ gets large?

Thank you! So if I understand correctly we're saying that $\int_1^T\frac{\zeta'(s)}{\zeta(s)}\chi(s)X^sdt=\sum_{n=1}^{\infty}\Lambda(n)(X/n)^c\int_1^T(X/n)^{it}\chi(c+it)dt$? In that case how does the $\ll T^{1/2-c}$ arise? Because in our case $c>1$ which means as $T$ gets large the integral gets small. Could you give some more explanation? I think the main part is to evaluate the integral and then combine with the sum, but I can't see a nice way of doing this because of the $\chi$ factor

@tomos could you explain please? I'm not sure why $t$ large is the main contribution. Can you outline how to bound the integral? I will look at Titchmarsh now, thank you!

@Haidara correct. I'm interested to see how it works in this case though, as in how one could estimate it in its current form

@Haidara $X$ is just a parameter, which is at most $T$. I'll add this to the post

Excellent answer - thank you! Just a couple follow up questions - (1) if our $\lambda=1$ the optimal bound we get using this method is $I(1)=O(1)$ -- with $\lambda=1$ is it ever possible to obtain a stronger bound (faster than polynomial decay)? (2) why does this integration by parts method not work for stationary phase?