We know that factorial of natural number n describes how many bijections there are from some set with k cardinality into itself. But what if cardinality of the set is non natural number? or what if the set is uncountable?
if factorial of non natural number n is $$n!=\int_{0}^{\infty}{e^{-x}x^{n}\:dx}$$ and n describes cardinality of some set, then waht is this result for the set?
On
What you actually use is the technique of analytic continuation, where a function defined on a subset of the complex plane (the integers here) is extended into an analytic function in a domain that is as large as possible; in this case the whole complex plane without the negative x-axis. But then, AFAIK, the meaning of $n!$ as , say,the number of orderings for {$1,2,..,n$} is lost when you extend beyond the natural numbers.
The integral-defined function's domain is the real numbers greater than $-1$. But non-natural real numbers can't even be cardinalities of sets -- what does it mean for a set to have $\frac{1}{2}$ an element in it? So, then, it's not meaningful to ask "how many permutations of a set containing $\frac{1}{2}$ elements are there?", because there isn't even any meaning to such a set.