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Let $x$ be a real number and $f(x)$ a real analytic function such that $f(x+1) = P(f(x),x)$ where $P$ is a given real polynomial. Express $f(x)$ as an integral from $0$ to $\infty$.

As an example we have the $\Gamma$ function.

I'm looking for the general solution.

I was thinking about the recursions used to compute integrals of type $\int$ $f(x)^k \mathrm{d}x$.

Also hypergeometric functions crossed my mind.

But I am stuck.

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How can you "let $f(x+1)=P(f(x),x)$", given that $f$ is already given? Do you mean "assume $f(x+1)=P(f(x),x)$?" – Thomas Andrews Nov 9 '12 at 19:20
Also it is totally unclear what you mean by "express $f$ as an integral transform". I doubt $f(x)=\int_0^{\infty}f(x)e^{-t}\,dt$ is what you are looking for. – fedja Nov 9 '12 at 19:34
@ThomasAndrews maybe i should connect those two sentenses with 'such that'. @ fedja do you know the integral transform that defines the gamma function ? I did not suggest putting f(x) in the integral itself ... maybe I should remove transform ? – mick Nov 9 '12 at 20:08
I edited. Hope it is better now. – mick Nov 9 '12 at 20:13
Do you have a reason to expect that this will be possible in this general form? – Lukas Geyer Nov 9 '12 at 20:14

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