Derivatives or Integrals of Factorials

calculusfactorial

I am wondering, why is it so hard to take the derivative or integral of a factorial function defined as $f(x) = x!$.

I know that approximations do exist, and it wouldn't be so hard to subject them to derivation or integration. I doubt that the reason is that factorials are defined only for whole numbers…

Is my thought right? What other things make this kind of a derivation seemingly hard (or impossible) to get ?

One thing I did is like this:
$$ (n+1)! – n! = n\times n!$$

This is a commonly known thing.

Now, derivatives are defined as (I am very much of a noob in calculus, even if I know a little bit of it):
$$\lim_{h\to 0}\dfrac{f(x+h) – f(x)}{h}$$
Taking $f(x) = x!$, when we try to take the derivative:
$$\lim_{h \to 0} \dfrac{(x+h)! – x!}h$$

In this case, simple substitution won't work, and since I don't know many other rules well, maybe I won't be able to work it out well.

But if I try to use Lagrange's mean value theorem to see if there is a place where the $f(x)$ here has a minima or a maxima (I only have a small knowledge about this) here, I'd have $f'(c) = \dfrac{a! – b!}{a-b}$ ($a$ and $b$ are distinct) and this can't be zero anywhere, since factorials are defined only for positive integers and $a = b$ means that the $f'(c)$ we get enters indeterminate forms.

Anything more that I can go through to understand why $\dfrac d{dx}{x!}$ is hard to evaluate ?

Best Answer

I am wondering, why is it so hard to take the derivative or integral of a factorial function

I doubt that the reason is that factorials are defined only for whole numbers.

That's exactly the reason. Calculus tools are made to handle functions defined on (most of) the real number line. That's their raison d'être. They can't really handle a function that is defined only on the integers, without either a massive reworking of calculus itself, or redefining the function to have a value everywhere.

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