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This is probably not a research level question but I honestly don't know how/where to look for techniques to reconstruct a function from its asymptotic expansion.

The expansion I want to know about occurred in connection with another question I asked before (this one); it is $$ \sum_{n\geqslant1}\frac{B_{2n}}{2n!}(-z)^{1-n}=\frac1{12}+\frac{z^{-1}}{120}+\frac{z^{-2}}{504}+\frac{z^{-3}}{1440}+\frac{z^{-4}}{3168}+\frac{691z^{-5}}{3931200}+... $$

My question is whether this expansion is familiar to anybody (it looks so "classical"!) or if not, whether there are general methods to obtain some representations (series, integrals) of a function from its asymptotic expansion.

(NB) Everything would be clear to me with (2n)! instead of 2n!

A later update:

Inspired by the comment of @IgorKhavkine I applied a slight modification of the Borel summation trick to the original asymptotic expansion and obtained \begin{multline*} \sum_{n\geqslant1}\frac{B_{2n}}{2n!}(-z)^{1-n}=\sum_{n\geqslant1}\frac{B_{2n}}{2n!}(-z)^{1-n}\frac{\int\limits_0^\infty t^{n-\frac12}e^{-t}dt}{\Gamma\left(n+\frac12\right)}\\=\sum_{n\geqslant1}\frac{B_{2n}4^n}{2(2n)!\sqrt\pi}(-z)^{1-n}\int\limits_0^\infty t^{n-\frac12}e^{-t}dt\\=\frac{-z}{2\sqrt\pi}\int\limits_0^\infty\frac1{\sqrt t}\sum_{n\geqslant1}\frac{B_{2n}(-4t/z)^n}{(2n)!}e^{-t}dt\\=\frac{-z}{2\sqrt\pi}\int\limits_0^\infty\frac1{\sqrt t}\left(\frac{\sqrt{-4t/z}}{e^{\sqrt{-4t/z}}-1}-1+\frac12\sqrt{-4t/z}\right)e^{-t}dt\\=\sqrt{\frac{-z}\pi}\int\limits_0^\infty\frac{e^{-t}}{e^{2\sqrt{-t/z}}-1}dt+\frac z2+\frac12\sqrt{\frac{-z}\pi} \end{multline*} At the moment I do not care so much about legitimacy of the above transformations; what matters to me is whether the obtained function coincides with the one I mentioned in a comment below. Moreover the function of $z$ I obtained in the end does indeed have the desired asymptotic expansion, except that I failed to find any information about the integral I have arrived at. Thus my question now becomes

Can the integral $$\int\limits_0^\infty\frac{e^{-t}}{e^{c\sqrt t}-1}dt$$ be evaluated in closed form using some known special functions?

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მამუკა ჯიბლაძე
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    There is the following asymptotic series: $f(t) = -\frac{1}{2}-\frac{1}{4\sqrt{t}}+\frac{\sqrt{t}}{2} \psi'(\sqrt{t}) = \sum_{n=1}^\infty \frac{1}{2} B_{2n} t^{-n}$ for $1/t\sim 0$. I think you should be able to generate the $n!$ in the denominators by applying some integral transform to $f(t)$, which escapes me at the moment. For the general question, the answer is No. One cannot recover a smooth function from its asymptotic expansion at a point. It becomes Maybe if you can make assumptions, like growth, analyticity in a complex wedge, etc. (see https://en.wikipedia.org/wiki/Divergent_series) – Igor Khavkine Feb 09 '14 at 13:34
  • Should have mentioned that $\psi(t)$ is the digamma function and $\psi'(t)$ is its derivative: http://functions.wolfram.com/GammaBetaErf/PolyGamma/ – Igor Khavkine Feb 09 '14 at 13:36
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    In general, an asymptotic expansion does not identify your function uniquely, unless you impose some additional restrictions. So you better explain where it comes from. – Alexandre Eremenko Feb 09 '14 at 15:05
  • @AlexandreEremenko I've encountered it when studying behaviour of $\tilde\theta(\tau):=\sum_{n\geqslant1}ne^{n^2\pi i\tau}$ under modular transformations: Poisson summation gives that the difference $$ \tilde\theta(-1/\tau)-\frac\tau{2\pi i}-\sqrt{\frac{\tau^3}i}\tilde\theta(\tau) $$ has this asymptotic expansion; and observing the contour plots of it seemingly it is holomorphic everywhere except that it has a branching pole at the origin. – მამუკა ჯიბლაძე Feb 09 '14 at 18:06
  • (Well more precisely one has to insert some $\pi$s and $i$s at appropriate places :) ) – მამუკა ჯიბლაძე Feb 09 '14 at 18:29
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    You can add any function which decays faster than any power, such as $e^{-z}$, and the asymptotic expansion remains the same. – Brendan McKay Feb 09 '14 at 23:48
  • @BrendanMcKay Yes it is disappointing... still I hope to identify the difference in my comment above among the functions with given asymptotics. – მამუკა ჯიბლაძე Feb 10 '14 at 06:04
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    In general, if you fix an angular sector $S$ of your point, the Borel-Ritt theorem gives you a surjection from holomorphic functions on $S$ with asymptotic expansions to formal power series. Flat functions on $S$ (i.e., those with asymptotic expansion zero) then form the kernel. There are some more refined versions of the theorem involving Gevrey growth conditions. – S. Carnahan Mar 01 '14 at 15:10

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