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<m<n. 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}

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