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An integral and a deravative of a simple factorial

  1. Jan 2, 2004 #1


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    how can you find the integral and the derevative of a simple factorial f(x)=x! (to find what f'(x) equals and what Sf(x)dx equals)? as i see it you have progressive multiplications, f(x)=x(x-1)(x-2)...*(x-k), which is the product of x-k where k=0 till infinity, should i take logarithms on both sides? if i have asked this before link me to the thread.
    thanks in advance.
  2. jcsd
  3. Jan 2, 2004 #2
    Well, you can't integrate or differentiate the traditional factorial function. You would have to take advantage of the fact that [itex]n!=\Gamma(n+1)[/itex] and operate on the gamma function instead.

    I can't help you with that though. I don't know a lot about the gamma function.
  4. Jan 2, 2004 #3


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    Integration and differentiation require, at least, that the function be defined on some interval of real numbers. The factorial function is only defined for non-negative integers. You can, as master_coda said, use the gamma function instead.
  5. Sep 20, 2009 #4
    you cannot differentiate a factorial...
    For a function to be differentiable, it has to be continuous. For discrete functions like x! the derivative does not exist.

    As for the integration goes, theoretically, it is possible to integrate x!.
    I am not sure though, that the gamma function approach will work. The result of the gamma function integration gamma(x+1) leads to x!. Hence replacing x! by its gamma forms leads to a double integral which will be more difficult to solve.
  6. Sep 21, 2009 #5
    The factorial function is not continuous, so you just use [itex](x-1)!=\Gamma(x)[/itex]. There is no known indefinite integral of the Gamma function. However, it does have a derivative in terms of itself and another function.
    [itex]\Gamma '(x)=\Gamma(x) \psi(x)[/itex] where [itex]\psi[/itex] is known as the digamma function.
  7. Sep 21, 2009 #6
  8. Sep 21, 2009 #7
    That can also be expressed with the polygamma function:

    [itex]\int log[\Gamma(z)]dx=\psi^{(-2)}(z)+C[/itex]
  9. Sep 21, 2009 #8
    Yes, but then the poygamma function of order minus two or smaller is nothing more than the (repeated) integral of Log(Gamma). The properties of these functions are not trivial. Barnes and others investigated the related Barnes G-function, otherwise Barnes would not have been bothered to do that. :smile:
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