Dirichlet Series and Asymptotic Expansions: $\tilde{f}(s)= \sum_{n=1}^{\infty} f(n) n^{-s}$

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Consider the Dirichlet series $\tilde{f}(s)= \sum_{n=1}^{\infty} f(n) n^{-s}$. In the page Zeta Function Regularization

I found a relation among an asymptotic expansion of $\tilde{f}(s)$ and an asymptotic expansion of $F(t)=\sum_{n=1}^{\infty}f(n)e^{-tn}$, that is if \begin{equation} F(t)=\frac{a_N}{t^N}+\frac{a_{N-1}}{t^{N-1}}+... \end{equation} then \begin{equation} \tilde{f}(s)=\frac{a_N}{s-N}+... \end{equation} Is it correct? Where could I find a rigorous treatment of such a subject?

Thank you very much in advance for your help.

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Just some basic facts.

Recall the classic integral representation for the Euler gamma function $$ \Gamma(s)=\int_0^{+\infty}x^{s-1} e^{-x}dx, \quad \Re s>0, \tag1 $$ then by the change of variables $x=nt$, where $n>0$, we obtain $$ \frac{1}{n^s}\Gamma(s)=\int_0^{+\infty}t^{s-1} e^{-nx}dx \tag2 $$ then formally $$ \begin{align} \Gamma(s)\tilde{f}(s)&= \sum_{n=1}^{\infty} f(n)\:n^{-s} \Gamma(s) \\\\ &=\sum_{n=1}^{\infty} f(n)\int_0^{+\infty}t^{s-1} e^{-nx}dx \\\\ &=\int_0^{+\infty}t^{s-1} \sum_{n=1}^{\infty} f(n)e^{-nx}dx \\\\ &=\int_0^{+\infty}t^{s-1} F(t)dt, \quad \Re s>0, \tag3\\\\ \end{align} $$ and the previous interchange between the sum and the integral will be justified on the set of absolute convergence of the Dirichlet series.

If you put $\displaystyle F(t):=\frac{a_N}{t^N}$ in $(3)$, then $$ "\Gamma(s)\tilde{f}(s)=a_N\int_0^{+\infty}t^{s-1} \frac{1}{t^N}dt=\frac{a_N}{s-N} " \tag4 $$ and you may see that the poles of $\tilde{f}$ in the s-plane correspond to divergent terms in the Laurent series of its exponential regularization $F$.

Here is a very interesting reference for these considerations: Flajolet Sedgewick book

Hoping to have helped you.