On the proof of Theorem 15.2 in Montgomery-Vaughan's Multiplicative number theory

Question

I am trying to understand the proof of Theorem 15.2 in the aforementioned book. In this proof, the authors seem to infer that, if $\psi(x) - x < x^{\Theta - \epsilon}$ for every $\epsilon > 0$ and sufficiently large $x$, then the function $f(s)=\int_{1} ^{\infty} (x^{\Theta - \epsilon} - \psi(x) + x)x^{-s-1} \mathrm{d}x$ is analytic in the half-plane $\Re(s)> \Theta-\epsilon$, where $\psi$ is the Chebyshev first function over prime powers.

But couldn't $\psi(x) -x$ be negative and large in absolute value, so that $f(s)$ would not be analytic in the said half-plane ? Or maybe I'm not understanding how they deduced the analyticity of $f(s)$ in that plane ?

Answer 1

It seems you are not using Lemma 15.1 correctly. For convenience I write $$\frac{1}{s - \Theta + \varepsilon} + \frac{\zeta'(s)}{s\zeta(s)} + \frac{1}{s-1} = \int_1^{+\infty} (x^{\Theta - \varepsilon} - \psi(x) + x) x^{-s-1} \, \mathrm{d}x. \quad (*)$$

Now the left-hand side of $(*)$ is analytic for real $s > \Theta - \varepsilon$ so Lemma 15.1 tells you that $\sigma_c \leq \Theta - \varepsilon$ (otherwise, since the RHS is analytic for $\mathfrak R(s) > \sigma_c$, the LHS, which coincides with the RHS for $\mathfrak{R}(s) > \sigma_c$ by the identity theorem, would have a singularity at $\sigma_c > \Theta - \varepsilon$).

Applying Lemma 15.1 once again yields that the LHS of $(*)$ is analytic for $\mathfrak R(s) > \Theta - \varepsilon$, which is a contradiction since $\zeta$ admits zeroes with $\mathfrak R(\rho) = \Theta - \varepsilon'$ for some $\varepsilon' < \varepsilon$.