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.



Estimates for $\sum_{n\leq x} d(n)^a$
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".


