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The Gamma function is defined by $$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\mathrm{d}t\tag{1}$$

and fulfills the property $$\Gamma(x+1)=x\Gamma(x)\tag{2}$$ I am wondering if following families of functions can be constructed that fulfill $$f(x+c)=xf(x)\tag{3}$$ $$g(x+c)=x^dg(x)\tag{4}$$ $$h(x+c)=k(x)h(x)\tag{5}$$

What I already found out

Loc & Tai modify eq.(1) by replacing $t^{x-1}\rightarrow t^{s(x-1)}$. The modified integral $w(x)$ fulfills the relation

$$w(x+1)=B(x)w(x)$$ where $B(x)$ is the Bernstein-Sato polynomial.

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  • $\begingroup$ Please check out the book On A Class of Incomplete Gamma Functions with Applications," by Chaudhry and Zubair, Chapman & Hall/CRC, 2002. The first chapter is 'Generalized Gamma Function.' $\endgroup$ Jul 12, 2021 at 19:21
  • $\begingroup$ @CyeWaldman The chapter of that book gives a different generalization of the Gamma function, and doesn't answer the question made here. $\endgroup$
    – jjagmath
    Jul 13, 2021 at 1:58

2 Answers 2

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Generically, any solutions to

$$f(x + 1) = x f(x)$$

can be given as $f(x) = \Gamma(x)\ \theta(x)$ for some 1-cyclic function $\theta$ such that $\theta(0) = 1$. Likewise, a similar $c$-cyclic deformation of each other equation is sufficient to generate all solutions once a "natural" representative has been found.

The question, then, is what is the most "natural" representative - particularly in the last case which subsumes the other two. This is actually quite tricky, because it seems while we can construct such for parts of it, it seems difficult to extend them to suitably general cases of functions. So this can only be a partial answer at best, and I cracked at this for a long time about a decade ago, but hadn't had tremendous success. What follows is a recap of how far I got.

The first thing to say with this regard is to note that simple argument scalings will convert your (4) into

$$f(x + 1) = h(x) f(x)$$

i.e. we don't need an arbitrary step of $c$. Next, observe that this is equivalent t the continuum product problem, i.e. generalizing

$$f(x) = \prod_{n=1}^{x} h(n)$$

to cases where the term count $x$ is a continuous, real number. And then by dropping a logarithm over it, we see that

$$\log f(x) = \sum_{n=1}^{x} \log(h(n))$$

and thus we relate this problem to the continuum sum - the question of generalizing

$$\sum_{n=1}^{x} a(n)$$

to cases where $x$ is a continuous real number.

And thus we ultimately need to consider the continuum sum problem - which is the most directly tractable form: the closest there seems to be to a "natural" way to do this is to consider analytic $a$ (or log-analytic), such that

$$a(x) = \sum_{k=0}^{\infty} b_k x^k$$

then we can use Faulhaber's formula which says that

$$\sum_{k=1}^{K} k^p = \frac{B_{p+1}(K+1) - B_p(1)}{p + 1}$$

where $B_p$ is the Bernoulli polynomial, so that we should define

$$\sum_{n=1}^{x} a(n) := \sum_{k=0}^{\infty} a_k \left(\frac{B_{k+1}(x+1) - B_k(1)}{k + 1}\right)$$

and this works for a variety of functions, but it also fails for many more. In particular, this sum diverges for even $a(x) = \log x$, meaning we cannot even recover the usual Gamma function, and I am not sure if there is any way to "analytically continue" it to arbitrary analytic functions - it quickly runs into being a complicated problem in "analytic continuation of operators" and I've not had luck before in finding an answer.

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For (4), assuming $c>0$, we have the solution $f(x) = (c^{x/c}\,\Gamma(x/c))^d$. Of course, (3) is the particular case of $d=1$.

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