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 $t^{ia} ?$