Gamma Function: Duplication, Multiplication, Reflection Formulae

$\displaystyle \Gamma(s) =\int_{0}^{\infty} x^{s-1} e^{-x} d x, \quad \mathfrak{R}(z)>0$

$\displaystyle \Gamma(z)=\lim _{m \rightarrow \infty} \frac{m ! m^{z}}{z(z+1) \cdots(z+m)}$

Weierstrass:

$\displaystyle \Gamma(z)=\frac{e^{-\gamma z}}{z} \prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right)^{-1} e^{z / n}$

Factorials (Shift Identity):

$\displaystyle \Gamma(s+1) = s\Gamma(s)$

Proof: Integration by parts!

$\displaystyle \begin{aligned} \Gamma(s+1) &=\int_{0}^{\infty} x^{s} e^{-x} d x \\ &=-x^{s} e^{-x}\Big|_{0}^{\infty}+\int_{0}^{\infty} s x^{s-1} e^{-x} d x \\ &=s \int_{0}^{\infty} x^{s-1} e^{-x} d x \end{aligned}$

$\displaystyle \Gamma(1)=1$

$\displaystyle \Gamma(n)=1 \cdot 2 \cdot 3 \cdots(n-1)=(n-1) !$

Relection Formula: (Euler)

$\displaystyle \Gamma(1-s) \Gamma(s)=\frac{\pi}{\sin (\pi s)}, \quad s \notin \mathbb{Z}$

Proof 1: Start with any product expansion fo ${\Gamma}$

$\displaystyle \Gamma(s)=\lim _{n \rightarrow \infty} \int_{0}^{n} t^{s-1}\left(1-\frac{t}{n}\right)^{n} d t=\lim _{n \rightarrow \infty} \frac{n !}{n^{n}} \prod_{k=0}^{n}(s+k)^{-1} n^{s+n}$

$\displaystyle \Gamma(s)\Gamma(1-s)=\Gamma(s) (-s)\Gamma(-s) =\lim _{n \rightarrow \infty} \frac{1}{s} \prod_{k=1}^{n} \frac{1}{1-\frac{s^{2}}{k^{2}}}$

But we have

$\displaystyle \sin (\pi z)=\pi z \prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)$

Note by differentiating twice, this is equivalent to showing

$\frac{\pi^{2}}{\sin ^{2} \pi z}=\frac{1}{z^{2}}+\sum_{n \geq 1}\left(\frac{1}{(z-n)^{2}}+\frac{1}{(z+n)^{2}}\right)$

and this can be established by noting that the difference of both sides is a bounded analytic function.

Therefore

$\displaystyle \Gamma(s)\Gamma(1-s)= \frac{\pi}{\sin (\pi s)}$

Proof 2:

$\displaystyle \Gamma(s) \Gamma(1-s)=\int_{0}^{\infty} \int_{0}^{\infty} u^{-s} v^{s-1} e^{-(u+v)} d u d v$

$\displaystyle x= u+v, y=\frac{v}{u}$

$\displaystyle \Gamma(s) \Gamma(1-s)=\int_{0}^{\infty} \int_{0}^{\infty} \frac{y^{s-1}}{1+y} e^{-x} d x d y=\int_{0}^{\infty} \frac{y^{s-1}}{1+y} d y=\frac{\pi}{\sin \pi s} \quad \text{(Contour integrate)}$

Beta function:

$\displaystyle \mathrm{B}(x, y)=\int_{0}^{1} t^{x-1}(1-t)^{y-1} d t=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}$

Proof:

$\displaystyle \begin{aligned} \Gamma(x) \Gamma(y) &=\int_{u=0}^{\infty} e^{-u} u^{x-1} d u \cdot \int_{v=0}^{\infty} e^{-v} v^{y-1} d v \\ &=\int_{v=0}^{\infty} \int_{u=0}^{\infty} e^{-u-v} u^{x-1} v^{y-1} d u d v \end{aligned}$

$\displaystyle u=zt, v=z(1-t)$

$\displaystyle \begin{aligned} \Gamma(x) \Gamma(y)& =\int_{z=0}^{\infty} \int_{t=0}^{1} e^{-z}(z t)^{x-1}(z(1-t))^{y-1} z d t d z \\ &=\int_{z=0}^{\infty} e^{-z} z^{x+y-1} d z \cdot \int_{t=0}^{1} t^{x-1}(1-t)^{y-1} d t \\ &=\Gamma(x+y) \mathrm{B}(x, y) \end{aligned}$

$\displaystyle \begin{aligned}\Gamma(s)&=\int_{0}^{\infty} x^{s-1} e^{-x} d x \\ &=2 \int_{0}^{\infty} u^{2s-1} e^{-u^{2}} d u\end{aligned}$

$\displaystyle \begin{aligned}\Gamma(\frac{1}{2}) &=2 \int_{0}^{\infty} e^{-u^{2}} d u\\ &=2\left( \int_{0}^{\infty} \int_{0}^{\infty} e^{-\left(u^{2}+v^{2}\right)} d v d u\right)^{1/2}\\ &=2 \left(\int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} e^{-r^{2}} r d r d \theta \right)^{1/2} =\sqrt{\pi} \end{aligned}$

Legendre Duplication Formula

$\displaystyle \Gamma(s) \Gamma\left(s+\frac{1}{2}\right)=2^{1-2 s} \sqrt{\pi} \Gamma(2 s)$

Proof 1: (Using integrals, change of variables)

$\displaystyle \begin{aligned} \frac{\Gamma(z) \Gamma(z)}{\Gamma(2 z)}=\mathrm{B}(z, z) &=\int_{0}^{1} u^{z-1}(1-u)^{z-1} d u\\ &=\int_{0}^{1}\left(\frac{1+x}{2}\right)^{z-1}\left(1-\frac{1+x}{2}\right)^{z-1}\left(\frac{1}{2} d x\right)\\ &=\frac{1}{2} \int_{0}^{1}\left(\frac{1+x}{2}\right)^{z-1}\left(\frac{1-x}{2}\right)^{z-1} d x\\ &=\frac{1}{2^{1+2(z-1)}} \int_{0}^{1}\left(1-x^{2}\right)^{z-1} d x \\ &=2^{1-2 z} \int_{0}^{1}\left(1-x^{2}\right)^{z-1} d x\\ &=2^{1-2 z} B\left(\frac{1}{2}, z\right)\\ &=2^{1-2 z} \frac{\Gamma\left(\frac{1}{2}\right) \Gamma(z)}{\Gamma\left(z+\frac{1}{2}\right)} \end{aligned}$

$\displaystyle \implies \Gamma(2 z)=(2 \pi)^{-1 / 2} 2^{2 z-1 / 2} \Gamma(z) \Gamma\left(z+\frac{1}{2}\right)$

Proof 2: (Product formula and Stirling)

$\displaystyle \Gamma(z)=\lim _{n \rightarrow \infty} \frac{n^{z} n !}{z(z+1)(z+2) \cdots(z+n)}$

$\displaystyle \begin{aligned} \Gamma(z) \Gamma(z+1 / 2)&=\lim _{n \rightarrow \infty} \left(\frac{n^{z} n !}{z(z+1) \cdots(z+n)}\right)\left(\frac{2^{n+1} n^{z+1 / 2} n !}{(2 z+1)(2 z+3) \cdots(2 z+(2 z+1))}\right)\\ &= \lim _{n \rightarrow \infty} \left(\frac{2^{2 n+2} (n !)^{2}n^{1 / 2}}{(2 n) ! 2^{2 z}(2 z+2 n+1)}\right) \left(\frac{(2 n)^{2 z}(2 n) !}{(2 z)(2 z+1) \cdots(2 z+2 n)}\right)\\ &= \lim _{n \rightarrow \infty} \left( \frac{2^{2 n+2} n^{1 / 2}(2 \pi n)\left(\frac{n}{e}\right)^{2 n}}{ (2z+2n+1) \sqrt{4 \pi n}\left(\frac{2 n}{e}\right)^{2 n} 2^{2 z}}\right)\Gamma(2z)\\ &=2^{1-2 z} \sqrt{\pi} \Gamma(2 z) \end{aligned}$

Gauss Multiplication Formula:

$\displaystyle \prod_{k=0}^{m-1} \Gamma\left(s+\frac{k}{m}\right)=(2 \pi)^{\frac{m-1}{2}} m^{\frac{1}{2}-m s} \Gamma(m s)$

Proof: (With Product Formula and Stiriling’s)

$\displaystyle \Gamma\left(z+\frac{k}{m}\right)=\lim _{n \rightarrow \infty} \frac{n ! n^{z+k / m-1}}{\left(z+\frac{k}{m}\right)\left(z+\frac{k}{m}+1\right) \cdots\left(z+\frac{k}{m}-1+n\right)}$

$\displaystyle \begin{aligned} \Gamma(z) \Gamma(z+1 / m) \Gamma(z+2/ m)\cdots \Gamma(z+\frac{m-1}{m}) &=\lim _{n \rightarrow \infty} \prod_{k=0}^{m-1} \frac{n ! n^{z+k / m-1}}{\left(z+\frac{k}{m}\right)\left(z+\frac{k}{m}+1\right) \cdots\left(z+\frac{k}{m}-1+n\right)}\\ &= \lim _{n \rightarrow \infty} \prod_{k=0}^{m-1} \frac{n ! m^n n^{z+k / m-1}}{ (m z+k)(m z+k+m) \cdots(m z+k-m+m n)}\\ &= \lim _{n \rightarrow \infty}\frac{n!^m m^{mn} n^{mz} n^{\frac{m+1}{2}} }{(m z)(m z+1) \cdots(mz-1+m n)}\\ &= \lim _{n \rightarrow \infty} \frac{n!^m m^{mn} n^{\frac{m+1}{2}}}{(mn)!m^{mz}} \left(\frac{(m n)^{m z}(m n) !}{(m z)(m z+1) \cdots(m z+m n)}\right)\\ &= (2 \pi)^{(m-1) / 2} m^{1 / 2-m z} \Gamma(m z)\\ \end{aligned}$

Proof 3: Consider

$\displaystyle F(x):=m^{x} \Gamma\left(\frac{x}{m}\right) \Gamma\left(\frac{x+1}{m}\right) \cdots \Gamma\left(\frac{x+m-1}{m}\right)$

Easy to see that ${F(x+1) =xF(x)}$
And each of the factors of ${F}$ is log-convex, so ${F}$ is log-convex. Therefore, by Bohr-Mollerup we are done once we show that ${F(1)=1.}$ (check this!)

Stirling’s Approximation:

$\displaystyle \Gamma(s+1) \sim \sqrt{2 \pi s}\left(\frac{s}{e}\right)^{s}$

$\displaystyle \Gamma(s+1)=\sqrt{2 \pi s}\left(\frac{s}{e}\right)^{s} \left(1+\frac{1}{12 s}+\frac{1}{288 s^{2}}-\frac{139}{51840 s^{3}}-\frac{571}{2488320 s^{4}} \ldots\right)$

Bohr-Mollerup theorem: ${\Gamma(s)}$ is only function ${F}$ on ${x>0}$ that satisfies
${F(1)=1,}$ and
${F(x+1)=x F(x)}$ for ${x>0}$ and
${F}$ is logarithmically convex.

Proof: Using ${F(n+1)=n!}$ , log-convexity gives the following inequalities for every ${n.}$

$\displaystyle \frac{(n-1)^{x}(n-1) !}{(x+n-1)(x+n-2) \cdots(x+1) x} \leq F(x) \leq \frac{n^{x} n !}{(x+n)(x+n-1) \cdots(x+1) x}\left(\frac{n+x}{n}\right)$
This shows that ${F(x)}$ is uniquely determined as the limit on both sides.

$\displaystyle \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$

$\displaystyle \frac{d}{d z} \log \Gamma(z)=\frac{\Gamma^{\prime}(z)}{\Gamma(z)}=-\gamma-\frac{1}{z}+\sum_{i=1}^{\infty}\left(\frac{1}{i}-\frac{1}{z+i}\right)$

$\displaystyle \text{where} \quad \gamma=\lim _{n \rightarrow \infty}\left(\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{n}-\log n\right)$

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