Find the inverse of the integral transform $ S_n\{f\}=\int_{0}^1f(x)x^{n}\,dx$

Question

Let $S$ denote the integral transform that maps every function $f$ integrable on (0,1) to a sequence $\left(S_n\{f\}\right)_{n\in\mathbb{N}}$: $$S:f\mapsto \left(S_n\{f\}\right)_{n\in\mathbb{N}}\text{ defined by: } S_n\{f\}=\int_{0}^1f(x)x^{n}dx$$

Note that it's quite similar to the Mellin transform, whose expression is $$M\{f\}(s)=\int_{0}^{\infty}f(x)x^{s-1}dx$$ the differences being that for $S$ the integration is over $(0,1)$ instead of the whole real axis, and the transformed function is taken only on the integers instead of the real numbers so that it defines a sequence and not a proper function.

My question is, wether it is possible to determine under what conditions $S$ is injective, and more importantly an expression if its inverse transform $S^{-1}$ that would enable us to recover $f$ from the sequence $S_n\{f\}$.

In the case of the Mellin transform, we know that under certain conditions the inverse Mellin transform is defined and thus we can recover the original function from its transformed with the formula:

$$M^{-1}\{\varphi\}(x)=\frac{1}{2\pi i}\int_{-i\infty}^{i\infty}\varphi(s)x^{-s}ds$$

Because of the similarities between $S$ and the Mellin transform, I thought this might help, but it didn't lead me anywhere so far.

In case $S^{-1}$ cannot be expressed explicitely, then I'd like to study the case where of a continuous variable $s$ instead of a discrete variable $n$ (i.e $S$ maps a function to another function, and not a sequence):

$$S'\{f\}(s)=\int_{0}^1f(x)x^{s}dx$$

Of course, the invertibility of $S$ and $S'$ are not equivalent.

Any suggestion?

Answer 1

Letting $x=e^{-t}$,

$$ S_n\{f\} = \int_{0}^{1}f(x)x^n\,dx = \int_{0}^{+\infty}f(e^{-t})e^{-t}e^{-nt}\,dt = \mathcal{L}(f(e^{-t})e^{-t})(n) $$ hence the inverse transform is achieved through

1. an application of $\mathcal{L}^{-1}$ (inverse Laplace transform)
2. a replacement of the variable with $-\log s$
3. a division by $s$.

If only the values of $S_n\{f\}$ for natural $n$s are known, one may exploit the fact that $$ f(x)\stackrel{L^2(0,1)}{=} \sum_{n\geq 0} \left((2n+1)\int_{0}^{1}f(x)P_n(2x-1)\,dx\right)P_n(2x-1)$$ where $P_n(2x-1)$ is a shifted Legendre polynomial, $$ P_n(2x-1)=(-1)^n\sum_{k=0}^{n}\binom{n}{k}\binom{n+k}{k}(-x)^k = \frac{1}{n!}\frac{d^n}{dx^n}(x^2-x)^n.$$