Notes on Whittaker & Watson, Chapter XII, part 1

Whittaker & Watson, A Course of Modern Analysis. Chapter XII, The Gamma Function.

12.1 The Weierstrass product definition of $1/\Gamma(z)$ $$ \frac{1}{\Gamma(z)}=ze^{\gamma z}\prod_{n=1}^\infty\left(1+\frac{z}{n}\right)e^{-\frac{z}{n}} $$ in which $\gamma=\lim_{m\to\infty}\frac{1}{1}+\frac{1}{2}+\cdots\frac{1}{m}-\log m$ is the Euler-Mascheroni constant.

The product is an entire function, i.e. analytic for all $z\in\mathbb{C}$.

12.11 Substituting the Euler-Mascheroni constant by its definition, we have Euler’s formula of $\Gamma(z)$ $$ \begin{align*} \Gamma(z) &= \frac{1}{z}\prod_{n=1}^\infty\left(1+\frac{1}{n}\right)^{-z}\left(1+\frac{z}{n}\right)^{-1}, & z\neq0,-1,-2,\cdots.\newline &=\lim_{n\to\infty}\frac{1\cdot2\cdot\cdots\cdot(n-1)}{z(z+1)\cdots(z+n-1)}n^z. \end{align*} $$ 12.12 The recurrence relation $\Gamma(z+1)=z\Gamma(z)$ and $\Gamma(n)=(n-1)!$ for positive integer $n$.

12.13 Evaluation of infinite product by the Gamma function.

For infinite product $$ \begin{align*} P&=\prod_{n=1}^\infty\frac{(n-a_1)(n-a_2)\cdots(n-a_k)}{(n-b_1)(n-b_2)\cdots(n-b_k)}\newline &=\prod_{n=1}^\infty\left(1-\frac{a_1}{n}\right)\cdots\left(1-\frac{a_k}{n}\right)\left(1-\frac{b_1}{n}\right)^{-1}\left(1-\frac{b_k}{n}\right)^{-1}\newline &=\prod_{n=1}^\infty\left(1-\frac{a_1+\cdots a_k-b_1-\cdots-b_k}{n}+O\left(\frac{1}{n^2}\right)\right), \end{align*} $$ the necessary condition of absolute convergence is that $a_1+\cdots a_k-b_1-\cdots-b_k=0$. We can therefore introduce a factor $\exp\{n^{-1}(a_1+\cdots a_k-b_1-\cdots-b_k)\}$, such that $$ \begin{align*} P&=\prod_{n=1}^\infty\frac{(1-\frac{a_1}{n})e^{\frac{a_1}{n}}\cdots(1-\frac{a_k}{n})e^{\frac{a_k}{n}}}{(1-\frac{b_1}{n})e^{\frac{b_1}{n}}\cdots(1-\frac{b_k}{n})e^{\frac{b_k}{n}}}\newline &=\frac{b_1\Gamma(-b_1)\cdots b_k\Gamma(-b_k)}{a_1\Gamma(-a_1)\cdots a_k\Gamma(-a_k)}\newline &=\frac{\Gamma(1-b_1)\cdots\Gamma(1-b_k)}{\Gamma(1-a_1)\cdots\Gamma(1-a_k)}. \end{align*} $$ 12.14 Reflection formula $$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}. $$ which follows from the Weierstrass product and the representation of the sinc function that $\frac{\sin\pi z}{\pi z}=\prod_{n=1}^{\infty}({1-\frac{z}{n}})({1+\frac{z}{n}})$.

[The infinite product representation of the sinc function is derived from the Weierstrass factorization of $\sin$ (§§7.5-6). Euler used this representation to solve the Basel problem, see Ed Sandifer, “Basel Problem with Integrals,” How Euler Did It, March 2004.]

12.15 Gauss and Legendre’s multiplication formula $$ \Gamma(z)\Gamma\left(z+\frac{1}{n}\right)\cdots\Gamma\left(z+\frac{n-1}{n}\right)=(2\pi)^{\frac{1}{2}(n-1)}n^{\frac{1}{2}-nz}\Gamma(nz). $$ Consider $$ \begin{align*} \phi(z)&=\frac{n^{nz}\Gamma(z)\Gamma\left(z+\frac{1}{n}\right)\cdots\Gamma\left(z+\frac{n-1}{n}\right)}{n\Gamma(nz)}\newline &=\frac{n^{nz}\prod_{r=0}^{n-1}\lim_{m\to\infty}\frac{1\cdot2\cdot\cdots\cdot(m-1)}{(z+\frac{r}{n})(z+\frac{r}{n}+1)\cdots(z+\frac{r}{n}+m-1)}m^{z+r/n}}{n\lim_{m\to\infty}\frac{1\cdot2\cdot\cdots\cdot(nm-1)}{nz(nz+1)\cdots(nz+nm-1)}(nm)^{nz}} & (\text{Euler’s formula})\newline & = \lim_{m\to\infty}\frac{\{(m-1)!\}^nm^{\frac{1}{2}(n-1)}n^{nm-1}}{(nm-1)!} \end{align*} $$ which is independent of $z$. Thus $\phi(z)=\phi(\frac{1}{n})$ $$ \begin{align*} \phi\left(\frac{1}{n}\right)&=\Gamma\left(\frac{1}{n}\right)\Gamma\left(\frac{2}{n}\right)\cdots\Gamma\left(\frac{n-1}{n}\right)\newline &=\left(\prod_{r=1}^{n-1}\Gamma\left(\frac{r}{n}\right)\Gamma\left(1-\frac{r}{n}\right)\right)^{\frac{1}{2}}\newline &=\left(\frac{\pi^{n-1}}{\prod_{r=1}^{n-1}\sin\frac{r\pi}{n}}\right)^{\frac{1}{2}} = (2\pi)^{\frac{1}{2}(n-1)}n^{-\frac{1}{2}}, \end{align*} $$ which gives the multiplication factor.

[An elementary proof of the fact that $\prod_{r=1}^{n-1}\sin\frac{r\pi}{n}=\frac{n}{2^{n-1}}$ without using the complex definition of $\sin z=\frac{e^{iz}-e^{-iz}}{2i}$ can be found in S.L. Loney’s Plane Trigonometry, §369, the knowledge of which is expected by the authors.]

When $n=2$, this becomes Legendre’s duplication formula $$ 2^{2z-1}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)=\pi^{\frac{1}{2}}\Gamma(2z). $$ 12.16 Expansions of the derivatives of $\log\Gamma$ $$ \frac{d}{dz}\log\Gamma(z)=-\gamma-\frac{1}{z}+z\sum_{n=1}^\infty\frac{1}{n(z+n)},\newline \frac{d^2}{dz^2}\log\Gamma(z+1)=\frac{1}{(z+1)^2}+\frac{1}{(z+2)^2}+\cdots. $$