Asymtotic for $\sum_{n \leq x} f(n)n^{ia}$ Let $f$ be an arithmetic function. If $$\lim_{x \rightarrow \infty}\frac{\sum_{n \leq x} f(n)}{x} = M \neq 0, $$ deduce the asymtotic formula for $$\sum_{n \leq x} f(n)n^{ia}$$ where $a$ is a real number constant.
I am not sure about what the question is asking. We have notation of asymptotic $f \sim g := \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = 1.$
So I have to find $g$ such that $$\sum_{n \leq x} f(n)n^{ia} \sim g.$$
Is it equivalent to $$\sum_{n \leq x} f(n)n^{ia} = h_1(x) + O(h_2(x)) ?$$
I usually see many theorem in analytic number theory write the terms in the way $$\sum_{n \leq x} f(x) = mainterm + O(something).$$
Is this thing also included as asymptotic formular ?
For the prove, I plan to use Abel summation formula (https://en.wikipedia.org/wiki/Abel%27s_summation_formula) to write 
$$\sum_{n \leq x} f(n)n^{ia} = h(x) + O(something)$$ by setting $$a_n = f(n), A(t) = \sum_{n\leq t}a_n, \phi(t) = t^{ia}.$$
But I am not sure if Abel formular valid for complex function, and more over what is the (complex) derivative of $t^{ia} ?$
 A: Abel's formula gives you
$$ \sum_{n \leq x} f(n) n^{ia} = A(x)x^{ia} - ia \int_{1}^{x} A(u) u^{ia-1} \, du. \tag{*}$$
Now using the assumption, let us write $A(u) = (M + \epsilon(u))u$ where $\epsilon(u) \to 0$ as $u \to \infty$. Then
\begin{align*}
\int_{1}^{x} A(u) u^{ia-1} \, du
&= \int_{1}^{x} (M + \epsilon(u)) u^{ia} \, du \\
&= \frac{M}{ia+1}(x^{ia+1} - 1) + \int_{1}^{x} \epsilon(u) u^{ia} \, du
\end{align*}
We claim that the last integral is $o(x)$ as $x \to \infty$. Indeed, one can adopt the idea of the Cesaro-mean to check that
$$ \frac{1}{x} \left|\int_{1}^{x} \epsilon(u) u^{ia} \, du\right| \leq \frac{1}{x} \int_{1}^{x} |\epsilon(u)| \, du \to 0 \quad \text{as} \quad x \to \infty. $$
Thus we have
$$\int_{1}^{x} A(u) u^{ia-1} \, du = \frac{M}{ia+1}x^{ia+1} + o(x). $$
Plugging this back to $\text{(1)}$ and using the relation $A(x) = Mx + o(x)$ gives
$$ \sum_{n \leq x} f(n) n^{ia} = \frac{M}{ia+1}x^{ia+1} + o(x), $$
which is essentially equivalent to
$$\sum_{n \leq x} f(n) n^{ia} \sim \frac{M}{ia+1}x^{ia+1}. $$
