Let's consider the following equation,
$$g(y) = \int_0^1 \frac{f(x)}{1-xy} \, dx$$
Let's assume that $g(y)$ is known. Is there a general way to determine $f(x)$? In other words, is there a way to invert the integral relation?
I know that in the case of a Laplace transform,
$$g(s) = \int_{-\infty}^{+\infty} e^{-st} f(t) dt$$ which is kind of related to my question, under certain conditions one can obtain $f(t)$ knowing $g(s)$ by doing
$$f(t) = \frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty} e^{st} g(s) ds$$
Is there anything similar for the integral I wrote above?
If there were, it would be quite useful for a project I'm working on. Any help would be really appreciated.
Using Jack D'Aurizio's hints from that previous question, we write $$ g(y) = \sum_{k=0}^\infty y^k \int_0^1 f(x) x^k\; dx $$ so if $g$ is analytic on a set containing $[0,1)$, with Maclaurin series $$ g(y) = \sum_{k=0}^\infty c_k y^k $$ you want to find $f$ so that $$ \int_0^1 f(x) x^k \; dx = c_k $$
The connection with Laplace transform is that $e^{-xs} = \sum_{k=0}^\infty (-xs)^k/k!$, so the Laplace transform of the restriction of $f$ to $[0,1]$ is $$\eqalign{ F(s) &= \int_0^1 f(x) \sum_{k=0}^\infty (-xs)^k/k!\; dx\cr &= \sum_{k=0}^\infty c_k (-s)^k/k!\cr}$$