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Let $a\neq 1$ be a positive constant and let $d(n)$ denote the number of divisors of $n.$

Can one obtain upper and lower bounds on $S_{a}(x)=\sum_{n\leq x} d(n)^a$?

I am particularly interested in estimates for $a\in(0,1)$ and $a=2$. A weak upper bound on the latter is $$S_2(x) \leq S_1(x)^2,$$ and can be used for the pair $(a,2a)$ in general, but surely more must be known.

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2 Answers 2

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One has $S_a(x) \sim C(a) x (\log x)^{2^a -1}$ where $$ C(a) = \Gamma(2^a)^{-1} \prod_p \left( 1 - \frac{1}{p} \right)^{2^a} \left( \sum_{k \geq 0} \frac{(k+1)^a}{p^k}\right). $$ This follows for example from standard tauberian theorems and from the fact that $\sum_{n \geq 1} d(n)^a n^{-s} = \zeta(s)^{2^a} F(s)$ where $F$ is an holomorphic function on the domain $\mathrm{Re}(s) > \frac{1}{2}$.

EDIT1: the estimate above is valid for any $a \geq 0$ and can be made uniform in $a$ when $a$ stays bounded (as in your question).

EDIT2: Using the Landau-Selberg-Delange method, it is possible to obtain a full asymptotic formula for the sum in question up to an error term of size $O(x/(\log x)^A)$ for any fixed $A$, even when $a$ is a complex number. Moreover, this estimate is uniform in $a$ when $|a|$ is bounded. See Tenenbaum's book "Introduction to Analytic and Probabilistic Number Theory".

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    $\begingroup$ In the special case $a=2$, the Euler product is $\prod_p ( 1 - 1/p^2) = \zeta(2)^{-1}= 6/\pi^2$ and the gamma factor is $\Gamma(4) = 6$, so the constant is $1/\pi^2$. Moreover one can probably use the formula $F(s) = \zeta(2s)^{-1}$ to get a better error term. $\endgroup$
    – Will Sawin
    Nov 17, 2017 at 9:51
  • $\begingroup$ How do you deal with Mellin inversion for non-meromorphic Dirichlet series, do you expand in real powers of $s-1$, then what ? $\endgroup$
    – reuns
    Nov 17, 2017 at 21:31
  • $\begingroup$ @reuns: I think for $a>0$ not an integer (and more generally for $a\in\mathbb{C}$ not an integer), the best tool is the Selberg-Delange method, see Tenenbaum's book for that. (And disregard my previous comment if you saw it, I deleted it quickly.) $\endgroup$
    – GH from MO
    Nov 18, 2017 at 1:04
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For elementary but more general and uniform upper bounds with the correct order of magnitude see these notes. In particular, consider (5) there in the special case $k=2$.

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