We know that the factorial can be extended to the whole complex plane except at negative integers and $0$ . But are there any theorems that allow us to do so ? . I know we can use the Identity theorem in complex analysis to extend functions on open sets but how does that work on the that case ? or is it just the integral representation that allows this extension ?

  • $\begingroup$ Probably the easiest way is the product representation of the $\Gamma$ function. $\endgroup$ Sep 2, 2013 at 10:21
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    $\begingroup$ Just to be sure, you're looking for a way to 'naturally' extended the factorial function to what we know will be the gamma function without first defining the gamma function and check it coincides (up to a translation) with the factorial function? $\endgroup$
    – Git Gud
    Sep 2, 2013 at 10:22
  • $\begingroup$ @GitGud , yes this is what I want . $\endgroup$ Sep 2, 2013 at 10:59

1 Answer 1


Since the factorial function is defined only on the natural numbers, it can't really be analytically continued. Rather, there are two steps involved here:

  1. Find an analytic function that coincides with the factorial function when plugging in a natural number.
  2. Analytically continue this function to the whole (almost) complex plane.

For the first step, $\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}\;\text{d}t$ does the job pretty well. For real inputs, it is a smooth curve that interpolates the factorial function in a nice way. Note that this is not the only analytic function that fulfills step 1 above. Just add any other analytic function which equals zero on the naturals, for example $\sin \pi z$, and you will get another one.

This function converges only for $\Re(s) > 0$. But it can be analytically continued to the whole complex plane, except at the negative integers. And this step is unique by use of the Identity theorem.

  • $\begingroup$ Thanks , this is helpful . $\endgroup$ Sep 2, 2013 at 11:00
  • $\begingroup$ @ZaidAlyafeai: Glad to help! $\endgroup$
    – Daniel R
    Sep 2, 2013 at 12:39

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