## Gamma Function, Sine Formula, and Gaussian Integral

This post is intended as a continuation of my last post, even if the chosen title does not suggests so. I want to prove some more results about the gamma function and in doing so, obtain an infinite product formula for $\sin x$, without using complex analysis. I’ll finish by using this infinite product (due to Euler) to show that

$\displaystyle\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$,

as opposed to the usual clever proof by polar integration.

Lemma. (de Moivre’s formula) Let $n\in \mathbb{Z}^{\geq 0}$. Then

$\displaystyle\left(\cos(\theta)+i\sin(\theta)\right)^{n}=\cos(n\theta)+i\sin(n\theta),\indent\forall\theta\in\mathbb{R}$

Proof. We prove de Movire’s formula by induction on $n$. The base case $n=0$ is obvious. Suppose the formula is true for some $n\in\mathbb{Z}^{\geq 1}$. Then

$\begin{array}{lcl}\displaystyle\left(\cos(\theta)+i\sin(\theta)\right)^{n+1}&=&\displaystyle\left(\cos(\theta)+i\sin(\theta)\right)^{n}\left(\cos(\theta)+i\sin(\theta)\right)^{n+1}\\&=&\displaystyle\left(\cos(n\theta)+i\sin(n\theta)\right)\left(\cos(\theta)+i\sin(\theta)\right)\\&=&\displaystyle\cos(n\theta)\cos(\theta)-\sin(n\theta)\sin(\theta)+i\left[\cos(n\theta)\sin(\theta)+\sin(n\theta)\cos(\theta)\right]\\&=&\displaystyle\cos((n+1)\theta)+i\sin((n+1)\theta)\end{array}$

where the last equality follows from an application of the sine and cosine angle sum formulas. $\Box$

We can use de Moivre’s formula to show that $\sin (2n+1)\theta$, for $n\in\mathbb{Z}$ is a polynomial in $\sin\theta$. Indeed,

$\begin{array}{lcl}\displaystyle\sin((2n+1)\theta)&=&\displaystyle\dfrac{-i}{2}\left[\cos((2n+1)\theta)+i\sin((2n+1)\theta)-\cos((2n+1)\theta)+i\sin((2n+1)\theta)\right]\\&=&\displaystyle\dfrac{-i}{2}\left[(\cos(\theta)+i\sin(\theta))^{2n+1}-\left(\cos(\theta)-i\sin(\theta)\right)^{2n+1}\right]\\&=&\displaystyle\dfrac{-i}{2}\sum_{k=0}^{2n+1}{{2n+1}\choose k}\left[(\cos \theta)^{k}(i\sin \theta)^{2n+1-k}-(\cos\theta)^{k}(-i\sin \theta)^{2n+1-k}\right]\\&=&\displaystyle\dfrac{-i}{2}\sum_{k=0}^{2n+1}{{2n+1}\choose k}(\cos\theta)^{k}\left[(i\sin\theta)^{2n+1-k}-(-i\sin\theta)^{2n+1-k}\right]\\&=&\displaystyle-i\sum_{k=0}^{n}{{2n+1}\choose 2k}(\cos\theta)^{2k}(-i\sin\theta)^{2n+1-2k}\\&=&\displaystyle\sum_{k=0}^{n}{{2n+1}\choose 2k}(-1)^{n+1-k}(\cos\theta)^{2k}(\sin\theta)^{2n+1-2k}\end{array}$

Making the substitution $\cos^{2}\theta=1-\sin^{2}\theta$ completes the proof of the claim. In fact, this polynomial is of degree $2n+1$. Indeed, $\sin((2n+1)\theta)$ has distinct roots (modulo $\pi$) at $\theta=\pm\frac{k\pi}{2n+1}$, for $k=0,\cdots,n$. Hence, the polynomial has roots at $\pm\sin(\frac{k\pi}{2n+1})$, for $k=0,\cdots,n$. So we can write $\sin(2n+1)\theta$ as the product

$\displaystyle\sin((2n+1)\theta)=c\sin(\theta)\prod_{k=1}^{n}\left(\sin^{2}(\theta)-\sin^{2}\left(\frac{k\pi}{2n+1}\right)\right)$

We now compute the constant $c$. If we divide both sides by $\sin\theta$ and compute the limit as $\theta \rightarrow 0$, we obtain

$\begin{array}{lcl}\displaystyle c\prod_{k=1}^{n}\left(-\sin^{2}\frac{k\pi}{2n+1}\right)=\lim_{\theta\rightarrow 0}\dfrac{\sin(2n+1)\theta}{\sin\theta}&=&\displaystyle\lim_{\theta\rightarrow 0}\dfrac{\sum_{k=0}^{\infty}\frac{(2n+1)^{k}\theta^{2k}}{(2k+1)!}}{\sum_{k=0}^{\infty}\frac{\theta^{2k}}{(2k+1)!}}\\&=&\displaystyle (2n+1)\end{array}$,

which implies that

$\displaystyle\sin(2n+1)\theta=(2n+1)\sin(\theta)\prod_{k=1}^{n}\left(1-\dfrac{\sin^{2}\theta}{\sin^{2}\left(\frac{k\pi}{2n+1}\right)}\right)$

Lemma. (Euler’s Sine Product)

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

Proof. Fix $x>0$ and positive integers $m$ and $n$ such that $x. The identity for $\sin (2n+1)\theta$ allows us to write

$\displaystyle\dfrac{\sin x}{(2n+1)\sin\frac{x}{2n+1}}=\prod_{k=1}^{n}\left(1-\dfrac{\sin^{2}\frac{x}{2n+1}}{\sin^{2}\frac{k\pi}{2n+1}}\right)$

Set $a_{k}:=1-\dfrac{\sin^{2}\frac{x}{2n+1}}{\sin^{2}\frac{k\pi}{2n+1}}$. By Jordan’s inequality, $\frac{2\theta}{\pi}<\sin\theta<\theta$ for $\theta\in(0,\frac{\pi}{2})$ and therefore

$\displaystyle\dfrac{4x^{2}}{\pi^{2}}=\left(\dfrac{\frac{2}{\pi}\cdot\frac{x}{2n+1}}{\frac{k\pi}{2n+1}}\right)^{2}<\dfrac{\sin^{2}\frac{x}{2n+1}}{\sin^{2}\frac{k\pi}{2n+1}}<\left(\dfrac{\frac{x}{2n+1}}{\frac{2}{\pi}\cdot\frac{k\pi}{2n+1}}\right)^{2}=\dfrac{x^{2}}{4k^{2}}$

Without loss of generality, we may assume that $m$ is sufficiently large so that $1-\frac{x^{2}}{4}\sum_{k=m+1}^{\infty}$.

$\displaystyle\sum_{k=m+1}^{n}\dfrac{1}{k^{2}}<\dfrac{m}{(m+1)^{2}}+\dfrac{m}{(2m+1)^{2}}+\cdots+\dfrac{m}{(jm+1)^{2}}$

where $n-m\leq jm$. Hence,

$\displaystyle1>a_{m+1}\cdots a_{n}>\prod_{k=1}^{n}\left(1-\dfrac{x^{2}}{4k^{2}}\right)>1-\dfrac{x^{2}}{4}\sum_{k=m+1}^{n}\dfrac{1}{k^{2}}>1-\dfrac{Cx^{2}}{4m}$

We obtain the estimate

$\displaystyle\left(1-\dfrac{Cx^{2}}{4m}\right)\prod_{k=1}^{n}\left(1-\dfrac{\sin^{2}\frac{x}{2n+1}}{\sin^{2}\frac{k\pi}{2n+1}}\right)<\dfrac{\sin x}{(2n+1)\sin\frac{x}{2n+1}}<\prod_{k=1}^{n}\left(1-\dfrac{\sin^{2}\frac{x}{2n+1}}{\sin^{2}\frac{k\pi}{2n+1}}\right)$

Since $(2n+1)\sin\frac{x}{2n+1}=(2n+1)\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2n+1)^{2k+1}(2k+1)!}$, we obtain that $\lim_{n\rightarrow\infty}(2n+1)\sin\frac{x}{2n+1}=x$. Since the series $\sum_{k=1}^{\infty}\frac{x^{2}}{4k^{2}}$ converges, we see that the infinite product converges absolutely and therefore we can evaluate the limit of the product by computing the limit of each factor. By the squeeze theorem,

$\displaystyle\left(1-\dfrac{Cx^{2}}{4m}\right)\prod_{k=1}^{\infty}\left(1-\dfrac{x^{2}}{k^{2}\pi^{2}}\right)\leq\dfrac{\sin x}{x}\leq\prod_{k=1}^{\infty}\left(1-\dfrac{x^{2}}{k^{2}\pi^{2}}\right)$

Letting $m\rightarrow\infty$, we conclude from another application of the squeeze theorem that

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

$\Box$

Using the identity $\Gamma(x)=\lim_{n\rightarrow\infty}\frac{n!n^{x}}{(n+x)(n-1+x)\cdots x}$, for $x>0$, we can write

$\begin{array}{lcl}\displaystyle\Gamma(x)\Gamma(1-x)&=&\displaystyle\lim_{n\rightarrow\infty}\dfrac{(n!)n^{x}(n!)n^{1-x}}{(n+x)\cdots x(n+1-x)\cdots (1-x)}\\&=&\displaystyle\lim_{n\rightarrow\infty}\left[\dfrac{n+1-x}{n}\cdot\prod_{k=1}^{n}\left(1-\dfrac{x^{2}}{k^{2}}\right)\right]^{-1}\\&=&\displaystyle\lim_{n\rightarrow\infty}\prod_{k=1}^{n}\left(1-\dfrac{x^{2}}{k^{2}}\right)^{-1}\\&=&\displaystyle\dfrac{\pi x}{\sin\pi x}\end{array}$

Since $\sin\frac{\pi}{2}=1$ and $\Gamma(\frac{1}{2})=\int_{-\infty}^{\infty}e^{-x^{2}}dx$, we conclude that

$\displaystyle\int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}$