Is there exist a function defined on the real or complex numbers that, restricted to the integers, is equal to the factorial?

Question

The title already contains the question, Basically I am wondering if it is possible to naturally define a function, continuous and differentiable, such that when restricted to the integers it is equal to the factorial.

Answer 1

Yes! This function is related to the gamma function - it turns out to be more convenient to define the gamma function with $\Gamma(n) = (n-1)!$.

This is defined on the set of all complex numbers without the non-positive integers (as expected, since there is no sensible way of having $(-1)!$ make sense with the usual factorial recurrence). It is also smooth, so is certainly continuous and differentiable.

Moreover, this satisfies a particular uniqueness property. There are many analytic (complex differentiable) functions interpolating the factorial function. For instance given any one, adding $k \sin m \pi x$ gives another valid solution. We can impose the additional constraint that $f$ must satisfy $f(1)=1$ and $$f(x+1) = x f(x)$$ for all positive reals $x$, which is stronger than the typical condition that $$f(x+1) = xf(x)$$ for all naturals $x$. But this still allows many solutions - given one we can multiply by, say, $e^{k \sin m \pi x}$ to give another valid solution.

To get uniqueness, the Bohr-Mollerup theorem says that if $f$ must additionally be logarithmically convex ($\log(f(x))$ is convex), then the only possible solution is $f(x) = \Gamma(x)$.

Answer 2

Yes, Gamma function is such a function.

$$\Gamma (z) = \int_0^\infty x^{z-1}e^{-x}dx$$

satisfies $$\Gamma (z+1)= z\Gamma (z)$$

Also, $$\Gamma (n)= (n-1)!$$