## 오늘의 계산 23 - Fourier Transform of the Extended Binomial Coefficient

2009.03.18 15:45

Today we are going to prove the unproven assertion in the previous posting 「오늘의 계산 16」, and also establish a proof of the observation in 「여러가지 잡담」.

Theorem. Let $\alpha$ be a complex number away from negative integers, and denote \begin{equation*} \binom{\alpha}{z} := \frac{\alpha!}{z!(\alpha-z)!} = \frac{\Gamma(\alpha+1)}{\Gamma(z+1)\Gamma(\alpha-z+1)} \end{equation*} the extended binomial coefficient. Then for $\Re(\alpha) > 0$ and $x \in \mathbb{R}$, we have $$\label{eqn:wts_1} \binom{\alpha}{x} = \sum_{n=0}^{\infty} \binom{\alpha}{n}\frac{\sin \pi(x-n)}{\pi(x-n)}$$ and $$\label{eqn:wts_2} \int_{\mathbb{R}}\binom{\alpha}{x} e^{-2\pi i \xi x} \; dx = \begin{cases} \big(1 + e^{-2\pi i \xi}\big)^{\alpha} & |x| < \frac{1}{2} \\ 0 & \text{otherwise}. \end{cases}$$
Proof. Let $I(\alpha, x)$ by $$\label{def:i} I(\alpha, x) = \int_{-\frac{1}{2}}^{\frac{1}{2}} \big(1 + e^{2\pi i \xi}\big)^{\alpha} e^{-2\pi i x \xi} \; d\xi.$$ We are going to prove that $I(\alpha, x)$ is equal to the right-hand side of \eqref{eqn:wts_1}, and then it is equal to the extended binomial coefficient $\binom{\alpha}{x}$. To prove the first claim, we make a heuristic calculation. By the extended binomial theorem, \begin{align*} I(\alpha, x) &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( \sum_{n=0}^{\infty} \binom{\alpha}{n} e^{2\pi n i \xi} \right)e^{-2\pi i x \xi} \; d\xi = \sum_{n=0}^{\infty} \binom{\alpha}{n} \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{2\pi (n-x) i \xi} \; d\xi \\ &= \sum_{n=0}^{\infty} \binom{\alpha}{n} \left[ \frac{e^{2\pi (n-x) i \xi}}{2\pi (n-x)} \right]_{-\frac{1}{2}}^{\frac{1}{2}} = \sum_{n=0}^{\infty} \binom{\alpha}{n}\frac{\sin \pi(x-n)}{\pi(x-n)} \end{align*} This calculation is justified once we prove that the integration and summation can be justified. This is accomplished if we make the estimate \begin{equation*} \left|\binom{\alpha}{n}\right| = O\left(\frac{1}{n^{1+\Re(\alpha)}}\right) \end{equation*} for $n = 0, 1, 2, \cdots$. Then Weierstrass $M$-test is sufficient for justification. Indeed, by Stirling's formula, \begin{align*} \left|\binom{\alpha}{n}\right| &= \frac{1}{n!}\left| \prod_{k=1}^{n} (\alpha - k + 1) \right| = \frac{1}{n!}\left| \prod_{k=1}^{n} (k - \alpha - 1) \right| = \frac{1}{n!}\left| \frac{\Gamma(n-\alpha)}{\Gamma(\alpha)} \right| \\ &\leq C \left| \frac{n^{-1/2} \left(\frac{n-\alpha}{e}\right)^{n-\alpha}}{n^{1/2} \left(\frac{n}{e}\right)^n} \right| \leq \frac{C}{n^{1+\Re(\alpha)}}. \end{align*} Now we prove $$\label{eqn:wts_3} I(\alpha, x) = \binom{\alpha}{x}.$$ Just observe that, once we prove it, both \eqref{eqn:wts_1} and \eqref{eqn:wts_2} are direct consequences of this relation. Indeed, \eqref{eqn:wts_2} follows from inverse Fourier formula applied to \eqref{def:i}. By integration by parts, \begin{align*} xI(\alpha, x) &= x \int_{-\frac{1}{2}}^{\frac{1}{2}} \big(1 + e^{2\pi i \xi}\big)^{\alpha} e^{-2\pi i x \xi} \; d\xi \\ &= \alpha \int_{-\frac{1}{2}}^{\frac{1}{2}} \big(1 + e^{2\pi i \xi}\big)^{\alpha-1} e^{-2\pi i (x-1) \xi} \; d\xi = \alpha I(\alpha-1, x-1) \end{align*} and by integration by substitution $\xi \mapsto -\xi$, \begin{align*} I(\alpha, x) &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \big(1 + e^{-2\pi i \xi}\big)^{\alpha} e^{2\pi i x \xi} \; d\xi \\ &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \big(1 + e^{-2\pi i \xi}\big)^{\alpha} e^{-2\pi i (\alpha-x) \xi} \; d\xi =I(\alpha, \alpha-x). \end{align*} Applying this relation, we have \begin{align*} (x+n)(\alpha-x+n)I(\alpha+2n,x+n) &= (x+n)(\alpha-x+n)I(\alpha+2n,\alpha-x+n) \\ &= (x+n)(\alpha+2n) I(\alpha+2n-1,\alpha-x+n-1) \\ &= (x+n)(\alpha+2n) I(\alpha+2n-1,x+n) \\ &= (\alpha+2n)(\alpha+2n-1) I(\alpha+2n-2,x+n-1). \end{align*} Successive application of this relation recursively, we have \begin{align*} I(\alpha, x) &= \left[ \prod_{k=1}^{n} \frac{(x+k)(\alpha-x+k)}{(\alpha+2k)(\alpha+2k-1)} \right] I(\alpha+2n, x+n) \\ &= \binom{\alpha}{x} \frac{(x+n)!(\alpha-x+n)!}{(\alpha+2n)!} I(\alpha+2n, x+n). \end{align*} From 「오늘의 계산 08」, we find that \begin{equation*} \lim_{n\to\infty} \frac{I(\alpha+2n, x+n)}{I(2n,n)} = \lim_{n\to\infty} \frac{\int_{-1/2}^{1/2} (1+e^{2\pi i \xi})^{\alpha} e^{-2\pi i x \xi} \cos^{2n}\pi \xi \; d\xi}{\int_{-1/2}^{1/2} \cos^{2n}\pi \xi \; d\xi} = 2^{\alpha}. \end{equation*} Also, from beta function identity, \begin{equation*} I(2n,n) = \int_{-\frac{1}{2}}^{\frac{1}{2}} (2 \cos \pi \xi)^{2n} \; d\xi = \frac{2^{2n}}{\pi} \beta\left(n+\frac{1}{2}, \frac{1}{2} \right) = \frac{2^{2n}}{\pi} \frac{\big( n-\frac{1}{2} \big)! \big( -\frac{1}{2} \big)!}{n!}. \end{equation*} Now, consider the situation where $n \to \infty$ and $a, b \in \mathbb{C}$ with $a \notin (-\infty, 0]$. Stirling's formula shows that \begin{align*} (an+b)! & \sim \sqrt{2\pi} (an+b)^{an+b+\frac{1}{2}} e^{-(an+b)} \\ & \sim \sqrt{2\pi} (an)^{an+b+\frac{1}{2}} \left( 1+\frac{b}{an} \right)^{an} e^{-(an+b)} \sim \sqrt{2\pi} (an)^{an+b+\frac{1}{2}} e^{-an}. \end{align*} Thus \begin{align*} & \frac{(x+n)!(\alpha-x+n)!}{(\alpha+2n)!} I(\alpha+2n, x+n) \\ & \sim \frac{2^{2n+\alpha}}{\sqrt{\pi}} \frac{(x+n)!(\alpha-x+n)!}{(\alpha+2n)!} \frac{\big( n-\frac{1}{2} \big)!}{n!} \\ & \sim \frac{2^{2n+\alpha}}{\sqrt{\pi}} \frac{\sqrt{2\pi} n^{n+x+\frac{1}{2}} e^{-n} \cdot \sqrt{2\pi} n^{n+\alpha-x+\frac{1}{2}} e^{-n}}{\sqrt{2\pi} (2n)^{2n+\alpha+\frac{1}{2}} e^{-2n}} \frac{\sqrt{2\pi} n^{n} e^{-n}}{\sqrt{2\pi} n^{n+\frac{1}{2}} e^{-n}} = 1. \end{align*} This proves \eqref{eqn:wts_3}, and therefore completes the proof.

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