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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 $\mathrm{Re}(s) > \frac{1}{2}$.

EDIT: 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).

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".

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 from 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 $\mathrm{Re}(s) > \frac{1}{2}$.

EDIT: 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).

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 $\mathrm{Re}(s) > \frac{1}{2}$.

EDIT: 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).

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 from 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 $\mathrm{Re}(s) > \frac{1}{2}$.

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 from 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 $\mathrm{Re}(s) > \frac{1}{2}$.

EDIT: 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).