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Well, the observation is that we can study the behavior of $\sum_{n \leq x} \frac{f(n)}{n^{s}}$ by Abel's summation formula. We have $$ \sum_{n \leq x} \frac{f(n)}{n^{s}} = \frac{1}{x^{s}} \sum_{n \leq x} f(n) + s \int_{1}^{x} \sum_{n \leq t} f(n) \frac{1}{t^{s +1}} dt$$. If we let x go to ininity and denote $\sum_{n \leq x} \frac{f(n)}{n^{s}}$ by $A(x)$, then we see that $\sum_{n = 1}^{\infty} \frac{f(n)}{n^{s}} = s \int_{1}^{\infty} \frac{A(t)}{t^{s + 1}} dt $ with a change of variable $t = e^{\xi}$ then this is $$ \sum_{n = 1}^{\infty} \frac{f(n)}{n^{s}} = s\int_{1}^{\infty} A(t)e^{-s\xi}d\xi$$
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Sun, 02 Aug 2020 22:27 GMT