Can Gamma Function be used to Integrate Factorials?

[Hyperreality]

By definite integral, gamma function can be defined as

\Gamma(z)= \int_{0}^{\infty} t^{z-1}e^{-t} dt

I've learned some properties of Gamma function but my lecturer didn't tell us the domain of Gamma function. (I'm assuming it is defined for all non-negative real numbers).

I thought of this problem a while ago:

We know that
\sum_{n=0}^\infty \frac{x^n}{n!} = \lim_{n\rightarrow\infty} (1+x/n)^n=e^1

My question is, is there a numerical solution to

\int_{0}^{\infty}\frac{1}{x!} dx

where x is an non-negative real number over a continuous interval in terms of gamma function?

 

[Zurtex]

I've tried looking this up but to no avail sorry. I think you will find that the Gamma function is definied for all complex numbers except negative real integers (but obviously depends on context of what you are using it for to what you definie it for). But something worth noting is that:

x! = \Gamma (x + 1) if and only if x \epsilon \mathbb{Z}, x \geq 0

So your question doesn't really make much sense, what you want to be asking is this:

\int_0^{\infty} \frac{1}{\Gamma (x + 1)}dx

Otherwise quite simply as x! is just a series of points:

\int_0^{\infty} \frac{1}{x!}dx = 0



P.S Err, also:

\sum_{n=0}^\infty \frac{x^n}{n!} \neq e^1

I think you will find that:

\sum_{n=0}^\infty \frac{x^n}{n!} = e^x

 

[flouran]

N! is just a series of points, where n is any whole number; whereas the Pi function can describe continuously. But there is a discrepancy; for instance, (1.5)! doesn't make sense in the traditional sense of the function, but it does according to the continuous Pi function. Thus, the integral of a factorial doesn't seem to make much sense. Besides, it wouldn't prove too useful anyways because in most mathematics, n! is presented as a constant and thus need not be integrated as a function over n.

 

[yasiru89]

You might try the Weierstrass product representation and have a go at

\int_0^{+\infty} \frac{1}{\Pi(t)}dt

, where \Pi(t) = \Gamma(t + 1)[/tex], but a nice answer probably won't result. You might try to get suitable Riemann sums at integer points on a general interval and try to bound the integral though.