Cauchy's integral formula is a powerful method to extract the $n$'th power series coefficient of an analytic function by evaluating a single complex integral. Is there any such analytic method to extract (ordinary) Dirichlet series coefficients? That is, assuming that a given function $f(s)$ admits a Dirichlet series expansion $$\sum_n a_n n^{-s},$$ is there any known method to compute a desired coefficient $a_n$?
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2$\begingroup$ Otherwise you can find the $a_n$ one by one from the asymptotic as $s \to +\infty$. I wonder if there is an analog of $c_n = \frac{F^{(n)}(0)}{n!} = \lim_{z \to 0} \frac{\sum_{k=0}^n {k \choose n} (-1)^{k-n} F(nz)}{z^n n!}$ $\endgroup$– reunsCommented Apr 25, 2019 at 21:34
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$\begingroup$ Possibly relevant: mathoverflow.net/questions/30975/… $\endgroup$– M.G.Commented Apr 26, 2019 at 12:40
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2 Answers
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Even for more general Dirichlet series $$f(z)=\sum_{0}^\infty a_n e^{-\lambda_nz}$$ there is the formula $$a_ne^{-\lambda_n\sigma}=\lim_{T\to\infty}\frac{1}{T}\int_{t_0}^Tf(\sigma+it)e^{\lambda_n it}dt,$$ where $t_0$ is arbitrary (real) and $\sigma>\sigma_u$, the abscissa of uniform convergence.
This formula determines both $\lambda_n$ and $a_n$: the RHS=0 when we integrate against $e^{i\lambda t}$ with $\lambda\neq \lambda_n$. The class of functions which can be represented by such a series is called (analytic) almost periodic functions (on a vertical line $\{s=\sigma+it:t\in R\}$). The "number-theoretic case" corresponds to $\lambda_n=n$.
Ref. S. Mandelbrojt, Series de Dirichlet, Paris, Gauthier-Villars, 1969. Dirichlet series with complex $\lambda_n$ have been also studied (by A. F. Leont'ev and his school).
answered Apr 25, 2019 at 21:21
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$\begingroup$ If we don't know the $\lambda_n$ coefficients, is there a way to obtain them as well, or the problem becomes way too undetermined? $\endgroup$– M.G.Commented Apr 26, 2019 at 8:30
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$\begingroup$ @M.G. Number-theoretic case is $\lambda_n=\log n$. In general, they are determined by the same formula I wrote: apply it with arbitrary $\lambda$, the RHS will be $0$ for all $\lambda$ except countably many $\lambda_n$. For the discussion of the class of functions $f$ which admit such a representation, see the book of Mandelbrojt that I mentioned. $\endgroup$ Commented Apr 26, 2019 at 11:26
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Yes. If $f(s)$ has a finite abscissa of absolute convergence $\sigma_a$, then $\forall \sigma > \sigma_a$: $$ \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^{T} f(\sigma+ it)n^{it} \mathrm{d}t = \frac{a_n}{n^{\sigma}}. $$ IIRC, the proof can be found in Apostol's book on Analytic Number Theory.
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1$\begingroup$ I think it holds for $\sigma > \sigma_c$ the absicssa of simple convergence because $f(s) = (s-s_0)\int_1^\infty (\sum_{m \le x} a_m m^{-s_0}) x^{-s+s_0-1}dx = O(s)$ so we can replace $f(\sigma+it)$ by $f(\sigma+i.) \ast \frac{ k}{\sqrt{2\pi}} e^{-t^2 k^2/2} = \sum_m a_m m^{-\sigma-it} e^{- \frac{\ln^2m}{2 k^2}}$ to make it absolutely convergent $\endgroup$– reunsCommented Apr 25, 2019 at 21:19