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I recently purchased H. M. Edwards' book entitled The Riemann Zeta Function. In the early pages of the volume, concerning the factorial function $\Gamma$, Edwards notes that

"Euler observed that $\Gamma(n)=\int_0^\infty e^{-x}x^{n-1}dx$."

My question is twofold:

  1. How does one "observe" such a formula? Surely, this does not merely come from an intuitive observation?
  2. How does one prove this formula, and more importantly, where does the techincal motivation for it come from?
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  • $\begingroup$ Funny, I was just reading the Wikipedia page on $\Gamma$ and the first section, "Motivation", gives a good explanation about the initial motivation to find an analytic continuation of the factorial, as well as explaining why the gamma function is what it is, out of an infinite number of possible interpolations. Probably what you are looking for. en.wikipedia.org/wiki/Gamma_function#Motivation $\endgroup$
    – andrepd
    Commented Nov 25, 2014 at 9:24
  • $\begingroup$ Question (2) could be considered as a definition of the Gamma function for real values. $\endgroup$
    – user_of_math
    Commented Nov 25, 2014 at 9:29
  • $\begingroup$ Very interesting maybe paywalled: link.springer.com/article/10.1007%2FBF00389433. $\endgroup$
    – Martín-Blas Pérez Pinilla
    Commented Nov 25, 2014 at 9:33
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    $\begingroup$ In contrast to what many people could say, intuition is very much dependent on experience. That was Euler, so that formula could come from his "intuition" - imagine you work with similar things, and see in parametric integrals their dynamics on the parameter, then you can say - Hey! this integral shall behave like $I(n+1) = n I(n)$. Yet, I guess in the current case a more likely scenario is Euler coming across this integral, and noticing that $I(n+1)$ can be expressed in terms of $I(n)$. That's just speculation, however, I guess the story shall be available in some math history book. $\endgroup$
    – SBF
    Commented Nov 25, 2014 at 9:43
  • $\begingroup$ @Martín-BlasPérezPinilla That looks like a great book. I might get it for Christman, thanks. $\endgroup$
    – Klangen
    Commented Nov 25, 2014 at 10:05

2 Answers 2

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If you have not seen integration by parts before, it is strongly related to the product rule of differentiation. $$\frac{d}{dx}x^ne^{-x}=nx^{n-1}e^{-x}-x^ne^{-x}\\ \left.x^ne^{-x}\right|_0^{\infty}=\int_0^{\infty}nx^{n-1}e^{-x}dx-\int_0^{\infty}x^ne^{-x}dx\\ \int_0^{\infty}x^ne^{-x}dx=n\int_0^{\infty}x^{n-1}e^{-x}dx$$ So the integral with $n$ is related to the integral with $n-1$; and by the same rule that $n!$ is.

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(1) According to the Wikipedia, the definition of $\Gamma$ by Euler was $$\Gamma(x)=\lim_{n\to\infty}\frac{n! n^x}{x(x+1)\cdots(x+n)}.$$ (2) See Baby Rudin, 8.17,8.18,8.19 or Equivalence of Definitions of Gamma Function.

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