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Let $f$ be an integrable real valued function defined on $[0,\infty)$. Let $$m_n=\int_0^\infty f(x)x^n \mathrm dx$$ be the $n^{th}$ moment, and suppose that all of these integrals converge absolutely. Are there conditions that can we impose on $f$ that would allow us to write $f$ explicitly in terms of its moments and certain simple functions.

This idea is similar to the fact that we can reconstruct sufficiently nice functions on $[0,1]$ from a sum of their Fourier coefficients and $e^{inx}$. Also, on $(-\infty,\infty)$ we can reconstruct $f$ from an integral over the real line.

The analogy for $[0,\infty)$ and $x^s$ is of course the Mellin transform, which has an inversion formula as a line integral in the complex plane.

My question is then: Can we impose nice enough (non-trivial) conditions on a class of functions so that we can invert the Mellin transform on the real line based only on its values at the positive integers?

Thanks!

Note: I am not asking about the moment problem and the requirements for uniqueness. (Such as Carleman's Condition etc..)

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    Surely you are aware of Post's formula http://en.wikipedia.org/wiki/Post%27s_inversion_formula to inverse a Laplace transform. To fit into your question, one can restrict the limit to a sequence of integer values.2011-04-13
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    If derivatives of Laplace transforms are not allowed, a solution when $f$ is continuous might be to approach it on $[0,T]$ by a Bernstein polynomial (the ones which yield a proof of Stone-Weierstrass approximation theorem based on the law of large numbers) of degree $N$ large enough and to use the explicit bounds of the error to let $T$ and $N$ go to infinity. But your functions $f$ are not continuous, are they?2011-04-13
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    @Didier: Thanks, this is very interesting. At the moment, I don't see how to make it fit my question exactly, or just Mellin Transforms, but I think I might find a way by reading the derivation. (Wikipedia gives a link to http://www.rose-hulman.edu/~bryan/invlap.pdf)2011-04-13
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    @Didier: I am interested in the most general case, but continuous should be nice enough. The function $f$ which motivated this question was actually analytic, and monotonic on $\mathbb{R}^+$. (But that is a bit too narrow a criterion!)2011-04-13
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    Eric: OK. Rereading myself, I fail to see how the "Stone-Weierstrass" comment applies, so you can probably forget it. Sorry about the noise.2011-04-13
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    I haven't gotten to the thing, but this [article](http://dx.doi.org/10.1007/BFb0085579) might have something you can use...2011-05-01
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    [Here](http://iopscience.iop.org/0266-5611/3/3/016) is another article you might want to look at.2011-05-09

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