Gamma reflection

Here's another cool identity from Euler. You can start by defining the factorial of a number, denoted n!n!, as that number multiplied by all the numbers below it. Mathematically, this means n!=n(n1)(n2)1n! = n \cdot (n-1) \cdot (n-2) \cdots 1. We can also write this recursively, as

n!=n(n1)!0!=1 n! = n \cdot (n-1)! \qquad 0! = 1

Taking nNn \in \mathbb{N} , the first few values indexed by nn are {1,2,6,24,125,750,5250,}\{1,2,6,24,125,750, 5250, \dots\} . Taking the factorial as a function, it seems that it can grow far more quickly than any exponential function. This can be easily checked by using the taylor series of f(x)=exf(x)=e^x:

en=k=0nkk! e^n = \displaystyle\sum_{k=0}^\infty \frac{n^k}{k!} en>nnn!n!>nnen=enln(n)n e^n > \frac{n^n}{n!} \quad \to \quad n! > \frac{n^n}{e^n} = e^{n \ln(n) - n}

This sets a lower bound on the growth of n!n!. An obvious upper bound is n!<nnn! < n^n, so we can roughly say that the factorial grows like nnn^n as nn approaches \infty. However, nnn^n doesn't fully satisfy the functional equation n!=n(n1)!n! = n(n-1)!, so there must be lower-order terms. These can be found with Stirling's approximation.

Here's a challenge: can we define the factorial for all real numbers? How about complex ones as well? It turns out we can, and this is where the gamma function can be introduced:

Γ(x)=0tx1etdt \Gamma(x) = \displaystyle\int_0^\infty t^{x-1}e^{-t}dt

This seemingly comes out of nowhere, but it relies on a following neat observation:

0tnetdt=0ntn1etdt \displaystyle\int_0^\infty t^{n}e^{-t}dt = \displaystyle\int_0^\infty n t^{n-1}e^{-t}dt

Thus it satisfies the same functional formula, Γ(x)=xΓ(x1) \Gamma(x) = x \Gamma(x-1) . Evaluating the full integral gives Γ(x)=(n1)!\Gamma(x) = (n-1)! whenever xx is an integer. We now have a factorial function, the Gamma function, which is continuously defined.

Due to the functional equation, this function blows up at all negative values. However, it stays finite for non-integer negative values. Thanks to the Weierstrass factorization theorem, we can factorize this function (or it's reciprocal) as a product of its poles (or zeroes) as:

1Γ(x)=xeγxn=1(1+x/n)ex/n \frac{1}{\Gamma(x)} = xe^{\gamma x} \displaystyle\prod_{n=1}^\infty (1 + x/n)e^{-x/n}

Where γ=limkn=1k1/nlnk0.57722\gamma= \lim_{k \to \infty} \sum_{n=1}^k 1/n - \ln k \approx 0.57722 . This will take a bit of work to justify. We can start with the peculiar identity

Γ(x)=limpp!px(x1)!(x+p)! \Gamma(x) = \lim_{p \to \infty} \frac{p! p^x (x-1)!}{(x+p)!}

Proof:

To prove this, we can expand (p+x)! (p+x)! as (x1)!(x)(x+1)(x+p) (x-1)! (x)(x+1)\cdots (x+p) and introduce a strange factor pxp!p^x p! to get

(x1)!=(p+x)!x(x+1)(x+p)=(p+x)!pxp!pxp!x(x+1)(x+p)(x-1)! = \frac{(p+x)!}{ x(x+1)\cdots (x+p)} = \frac{(p+x)!}{p^x p!} \frac{p^x p!}{ x(x+1)\cdots (x+p)}

The leftmost fraction expands to

(p+x)!pxp!=p!p!(p+1)(p+2)(p+x)px=(1+1/p)(1+2/p)(1+x/p) \frac{(p+x)!}{p^x p!} = \frac{p!}{p!} \frac{(p+1)(p+2)\cdots(p+x)}{p^x} = (1 + 1/p)(1 + 2/p) \cdots (1 + x/p)

Which approaches 11 as p p \to \infty. This must mean that

(x1)!=(p+x)!pxp!pxp!x(x+1)(x+n)=limppxp!x(x+1)(x+p) (x-1)! = \frac{(p+x)!}{p^x p!} \frac{p^x p!}{ x(x+1)\cdots (x+n)} = \lim_{p \to \infty} \frac{p^x p!}{ x(x+1)\cdots (x+p)}

Or Γ(x)=limpp!px(x1)!(x+p)! \Gamma(x) = \lim_{p \to \infty} \frac{p! p^x (x-1)!}{(x+p)!}

It takes some more work to prove that this works with any argument (it's definitely a challenging exercise). I'll skip the details here though. From here we can find the Weierstrass factorization

Γ(x)=limpexlnpx(1+x)(1+x/2)(1+x/p) \Gamma(x) = \lim_{p \to \infty} \frac{e^{x \ln p}}{x(1+x)(1+x/2) \cdots (1+x/p)} =limpex(lnp11/21/31/p)xn=1p11+x/nex/n = \lim_{p \to \infty} \frac{e^{x (\ln p - 1 - 1/2 - 1/3 - \dots - 1/p)}}{x} \displaystyle\prod_{n=1}^p \frac{1}{1 + x/n} e^{x/n} Γ(x)=eγxxn=111+x/nex/n \Gamma(x) = \frac{e^{- \gamma x}}{x} \displaystyle\prod_{n=1}^\infty \frac{1}{1 + x/n} e^{x/n}

This leads to a really interesting functional equation for the gamma function, known as Euler's reflection formula. The Weierstrass factorization of sin(πx)\sin (\pi x) is given by:

sin(πx)=πxnZ(1x/n)=πx(n=1(1x/n))(n=1(1+x/n)) \sin (\pi x) = \pi x \displaystyle\prod_{n \in \mathbb{Z}} (1-x/n) = \pi x \left( \displaystyle\prod_{n =1}^\infty (1 - x/n) \right) \left( \displaystyle\prod_{n =1}^\infty (1 + x/n)\right)

The big terms suggest that we can relate this with Γ(x)\Gamma(x) in some way. We can derive:

n=1(1+x/n)=ϵ(x)xeγxΓ(x) \displaystyle\prod_{n =1}^\infty (1 + x/n) = \frac{\epsilon(x)}{xe^{\gamma x} \Gamma (x)} n=1(1x/n)=eγxxϵ(x)Γ(x) \displaystyle\prod_{n =1}^\infty (1 - x/n) = \frac{-e^{\gamma x}}{x \epsilon (x) \Gamma (-x)}

Where ϵ(x)=n=1ex/n \epsilon(x) = \displaystyle\prod_{n =1}^\infty e^{-x/n} . Replacing the separated terms in the Weierstrass factorization of sin(πx)\sin (\pi x) yields:

sin(πx)=πx(ϵ(x)xeγxΓ(x))(eγxxϵ(x)Γ(x))=πxΓ(x)Γ(x) \sin (\pi x) = \pi x \left( \frac{\epsilon(x)}{xe^{\gamma x} \Gamma (x)} \right) \left( \frac{-e^{\gamma x}}{x \epsilon (x) \Gamma (-x)} \right) = \frac{- \pi}{x \Gamma(x) \Gamma(-x)}

Appealing to the functional equation Γ(x+1)=xΓ(x)\Gamma(x+1) = x\Gamma(x) leads us to simplify xΓ(x)x \Gamma(-x) as Γ(1x)-\Gamma(1-x), giving us the full reflection identity:

sin(πx)=πΓ(x)Γ(1x) \large \sin (\pi x) = \frac{\pi}{\Gamma(x) \Gamma(1-x)}

Note by Levi Adam Walker
1 week, 4 days ago

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