# Why is Euler's Gamma function the "ideal" expansion of the factorial function to the reals?

There are whole lots (an infinitude) of smooth features that accompany f (n) =n! on the integers. Exists a straightforward reason that Euler's Gamma function $\Gamma (z) = \int_0^\infty t^{z-1} e^t dt$ is "ideal"? Specifically, I'm seeking factors that I can clarify to first-year calculus pupils.

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2019-05-06 23:16:37
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Wielandt's theory claims that the gamma - function is the only function $f$ that pleases the buildings :

• $f(1)=1$
• $f(z+1)=zf(z)$ for all $z>0$
• $f(z)$ is analytic for $\operatorname{Re}z>0$
• $f(z)$ is bounded for $1\leq \operatorname{Re}z\leq 2$

(See additionally the relevant MathOverflow string Importance of Log Convexity of the Gamma Function, where I learnt more about the above theory.)

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2019-05-08 20:58:03
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This is a comment uploaded as a solution for absence of online reputation.

Adhering To Qiaochu Yuan, the gamma function turns up in the useful formula of the zeta function as the consider the Euler item representing the "prime at infinity", and also it takes place there as the Mellin change of some gaussian function. (Gaussian features take place subsequently as eigenvectors of the Fourier change.)

This goes to the very least as old as Tate's thesis, and also a feasible reference is Weil's Basic Number Theory .

EDIT. Artin was just one of the first individuals to promote the log - convexity building of the gamma function (see his publication on the function concerned), as well as additionally probably the first mathematician to totally recognize this Euler - variable - at - infinity facet of the very same function (he was Tate's thesis expert). I assumed his name needed to be stated in a conversation concerning the gamma function.

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2019-05-08 20:48:45
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For whatever factor, Nature (through which I suggest integrals) appears to favor the Gamma function as the "proper" alternative to the factorial in numerous integrals, which appears ahead essentially from its indispensable definition. As an example, for non - adverse integers $a, b$, it's not tough to show (and also there's an actually wonderful probabilistic argument) that

$\displaystyle \int_{0}^1 t^a (1 - t)^b \, dt = \frac{a! b!}{(a+b+1)!}.$

For (non - adverse?) actual values of $a$ and also $b$ the proper generalization is

$\displaystyle \int_0^1 t^a (1 - t)^b \, dt = \frac{\Gamma(a+1) \Gamma(b+1)}{\Gamma(a+b+2)}.$

And, certainly, integrals are necessary, so the Gamma function have to additionally be necessary. As an example, the Gamma function shows up in the basic formula for the volume of an n-sphere. Yet the outcome that, for me, actually compels us to take the Gamma function seriously is its look in the functional equation for the Riemann zeta function.

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2019-05-08 19:48:12
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Actually there are various other (much less - regularly) made use of expansions to the factorial, with various buildings from the gamma function which might be preferable in some contexts.

Euler's Gamma Function

Luschny's factorial function

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2019-05-08 19:33:46
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The Bohr–Mollerup theorem reveals that the gamma function is the only function that pleases the buildings

• $f(1)=1$ ;
• $f(x+1)=xf(x)$ for every single $x\geq 0$ ;
• $\log f$ is a convex function.

The problem of log - convexity is specifically vital when one intends to confirm numerous inequalities for the gamma function.

Incidentally, the gamma function is not the only meromorphic function pleasing $$f(z+1)=z f(z),\qquad f(1)=1,$$ without absolutely nos and also no posts apart from the factors $z=-n$ , $n=0,1,2\dots$ . There is an entire family members of such features, which, as a whole, have the kind $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{m=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is a whole function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\quad k\in\mathbb Z,$$ ( $\gamma$ is Euler's constant). The gamma function represents the most basic selection $g(z)=\gamma z$ .

Modify : dealt with index in the item.

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2019-05-08 19:08:01
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