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In the classical divisor problems, for $k\geq 2$, $\alpha_k$ usually denotes the infimum of real numbers $\sigma<1$ such that $$\Delta_k(x)=\sum_{n\leq x}d_k(n)-\textrm{Res}\left(\frac{\zeta^k(z)x^z}{z},z=1\right)=O(x^{\sigma})$$ as $x\rightarrow\infty$.

As often happens in analytic number theory, one might expect that $\alpha_k$ is also the infimum of $\sigma<1$ such that the integral $$\lim_{T\rightarrow\infty}\int_{\sigma-iT}^{\sigma+iT}\frac{\zeta^k(s)x^sds}{s}$$ exists for all $x>0$.

However, it appears to me that this equivalence is not so easily proved, nor can I find a proof in the literature. Therefore, I would like to ask if this is actually a known theorem or, if not, whether the difference has been been investigated and in which paper(s)?

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It is known (see Titchmarsh Chapter 12) that if you define $\gamma_k$ the lower bound of $\sigma > 0, \int_{-\infty}^{\infty}\frac{|\zeta(\sigma+it)|^{2k}}{|\sigma+it|^2}dt < \infty$, then $\frac{k-1}{2k}\le \gamma_k=\beta_k \le \alpha_k$, where $\beta_k$ is the usual mean bound of $\Delta_k$ ( lowest bound of orders for which $\frac{1}{x}\int_{0}^{x}{\Delta_k(y)^2}dy=O_{\epsilon}(x^{2\beta_k+\epsilon})$) and the integral above in the question converges to $\Delta_k(x)$ for $\gamma_k< \gamma <1$.

Since it is known that for example $\gamma_2=\frac{1}{4}, \gamma_3=\frac{1}{3}, \gamma_4=\frac{3}{8}$ (best possible, first two results appearing in Titchmarsh, last in Ivic, chaper 13), while the known respective values for $\alpha$ are still far away from those, even in the simplest cases $k=2,3,4$ the equivalence required is far from being proven and it is equivalent to fully solving the Dirichlet Divisor problem for $k=2,3,4$.

In general, obviously, even less is known about $\gamma_k, \alpha_k$ so the question is on par with Lindelof in many ways (as Lindelof is equivalent to any and all of $\alpha_k \le \frac{1}{2}, \gamma_k \le \frac{1}{2}, \gamma_k = \frac{k-1}{2k}$)

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  • $\begingroup$ Thank you, Conrad. I am aware of the known results on $\alpha_k$ and $\gamma_k$ and their ordering, but are you saying that it is a known theorem that the integral in my question converges for $\sigma>\gamma_k$? If so, that would indeed answer my question. $\endgroup$ – Kevin Smith Apr 24 '19 at 23:51
  • $\begingroup$ Yes, it is proven in Titchmarsh, chapter 12 as noted, though the proof is not hard since it uses the uniform convergence to zero of $\frac{\zeta(s)^k}{s}, \gamma +\epsilon \le \sigma <1$ (at infinity ) which follows from the finitness of the integral defining $\gamma$ and some standard stuff plus Cauchy and the convergence of the integral near $1$ $\endgroup$ – Conrad Apr 25 '19 at 0:50

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