Is there a general formula to solve factorial-like recurrence relations?

Question

It is pretty straightforward to show why Gauss' $\Pi$- function, $$\Pi(y)=\int_0^\infty x^y e^{-x}dx,$$ satisfies the recurrence relation $$\Pi(y)=y\Pi(y-1).$$ Now, I am interested whether there is a general method to obtain a solution, like the integral above, to the recurrence relation $$G(y) = f(y)G(y-1)$$ where $f$ can be chosen arbitrarily.

Answer 1

Well, a function $G$ satisfying this equation can look like anything on $[0,1[$, then it has to be extended in an unique way so as to satisfy the functional equation.

Edit: actually there is uniqueness if you are looking for solutions defined on $\mathbb R^+$ or if you suppose $f$ does not vanish, as otherwise there may be several choice for $G(y-1)$.

Answer 2

Let us consider an integrand of the form $\phi(x,y)\,e^{-x}$.

By parts,

$$\int_0^\infty\phi(x,y)\,e^{-x}dx=-\left.\phi(x,y)\,e^{-x}\right|_0^\infty+\int_0^\infty\dfrac{\partial\phi(x,y)}{\partial x}\,e^{-x}dx$$

and you want this to be

$$\int_0^\infty f(y)\,\phi(x,y-1)\,e^{-x}dx.$$

We can ignore the first term by using $\phi(x,y)-\phi(0,y)$ instead of $\phi(x,y)$, and by assuming that the growth rate in $x$ does not exceed $e^x$. Then we end-up with the equation

$$\dfrac{\partial\phi(x,y)}{\partial x}=f(y)\,\phi(x,y-1),$$ which unfortunately is of a mixed functional-differential type.