I have been trying to read Theorem 2.7 of Montgomery-Vaughan's "Multiplicative number theory" volume 1, and there is an issue I have run into: in the proof of subpart (e), when they write $$\delta \int_1^\infty \frac{\, dx}{x^{1+\delta}} \sum_{n \leq x} \frac{\Lambda(n)}{n \log n} = \sum_{n \geq 1} \frac{\Lambda(n)}{n^{1+\delta} \log n} = \log \zeta(1+\delta),$$ they refer to the Corollaries 1.3, 1.11 and 1.13. From what I can see, they want to apply Corollary 1.3 to the Dirichlet Series corresponding to the sequence $(a_n)_{n \geq 1}$ given by $a_n:= \Lambda(n) / (n \log n)$. This Dirichlet series is $$\sum_{n \geq 1} \frac{\Lambda(n)}{n^{1+s}\log n} = \log \zeta(s+1) \hspace{5mm} \text{ for } \Re(s)>0$$ and is clearly seen to have abscissa of absolute convergence (hence also abscissa of convergence) at most $0$ (and I think equality holds, which is basically the root of my problem). However in Corollary 1.3 (in the part they seem to be using in order to get the aforementioned identity) applies only when the abscissa of convergence is strictly negative. Even using Corollary 1.13 (which is the result stating the meromorphic continuation of the Riemann Zeta to the half plane $\sigma>0$ with simple pole of residue $1$ at $s=1$) doesn't seem to be useful - as $\log \zeta(s+1)$ then has a singularity at $s=0$, so it seems to give a result of the same strength as the one given by our absolute convergence argument above. I would really like to know what I am (probably unnecessarily) tearing my hair out over. Thanks.
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2$\begingroup$ the theorem holds also for $\sigma >0$, you forgot to read the rest of it xD (1.10) of the theorem holds always, only the boundedness of the coefficients change $\endgroup$– tomosOct 7, 2020 at 23:45
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1$\begingroup$ Facepalm, wow thanks so much. Good timing of your answer too: realization dawns upon me as I decide to get the sleep I've been denying myself for too long. :D $\endgroup$– AK12N1Oct 8, 2020 at 1:04
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