Explicitly reconstructing a function from its moments 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..)
 A: I don't know why you restrict your expression to positive x, because if you let your integral range from negative to positive infinity, the sum of the moments seems to have a pretty simple interpretation. Just take the Fourier Transform of your expression with f(x) and you get the definition of the Taylor series for F(w). (Remember that the Fourier Transforms of the powers of x are just the derivatives of the delta function.) In other words, the functions that are expressible in terms of the sum of their moments are just the functions whose Fourier Transforms are analytical.
A: (Partial answer) You might try looking at Newton interpolation and then an inverse Mellin transform on the result:
With $\bigtriangledown^{s-1}_n=\sum_{n=0}^{\infty}(-1)^n \binom{s-1}{n}[...]$,  formally
$g(s)=\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{m_j}{j!}$
$=\int_0^\infty f(x)\bigtriangledown^{s-1}_{n} \bigtriangledown^{n}_{j} \frac{x^j}{j!} dx$
$=\int_0^\infty f(x) \frac{x^{s-1}}{(s-1)!}dx$= a modified Mellin transform, and consequently,
$m_j=j!g(j+1)=\int_0^\infty f(x) x^jdx$.
Then performing the modified inverse Mellin transform
$f(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{sin(\pi s)} g(s) \frac{x^{-s}}{(-s)!} ds$.
You could then investigate what classes of functions allow these maneuvers.
Trivial example:
With $m_j=a^j$, then $g(s)=\frac{a^{s-1}}{(s-1)!}$ analytically continued from the Newton interpolation to all $s$, and
$f(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty}  a^{s-1} x^{-s} ds=\delta(x/a-1)/a= \delta(x-a)$.
Another example:
Let $U(\alpha,\beta,z)$ be the confluent hypergeometric function of the second kind.
With $m_j=j!U(j+1,j+1+b,a)$, then $g(s)=U(s,s+b,a)$ analytically continued from the Newton interpolation to all $s$, and
$f(x)=\displaystyle\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+ i\infty} \frac{\pi}{sin(\pi s)} U(s,s+b,a) \frac{x^{-s}}{(-s)!} ds=e^{-ax}(1+x)^{b-1}$.

Edit
(Perhaps  more to the point of your question.) The mathoverflow link in the comment below illustrates Ramanujan's Master Theorem / Formula, which gives f(x) as a Taylor or power series in terms of the moments for a certain class of functions. See also Chaudry and Quadir "Extension of Hardy's Class for Ramanujan's Interpolation Formula and Master Theorem with Applications" and Amdeberhan, Espinosa, Gonzalez, Harrison, Moll, and Straub "Ramanujan's Master Theorem."
