Evaluation of $\lim_{x \to k\pi} \frac{\sin(x)}{x(x-k\pi)}$ from the Weierstrass product expansion of $\sin(x)$

Question

Consider the Weierstrass infinite product expansion of the $\sin(x)$ function: $$ \tag 1 \frac{\sin(x)}{x} = \prod_{n=0}^\infty \left( 1 - \frac{x^2}{n^2 \pi^2} \right) = \prod_{n\neq 0} \left( 1 - \frac{x}{n \pi} \right). $$ Moreover, it is easily shown that, given an integer $k \neq 0$, $$ \tag 2 \lim_{x \to k \pi} \frac{\sin(x)}{x(x-k\pi)} = \frac{(-1)^k}{k\pi}. $$ I wanted to obtain (2) from the expansion (1). However, I seem to get a wrong result, and I'm not finding the mistake in my reasoning.

Starting from (1), I write $$ \tag 3 \frac{\sin(x)}{x(x-k\pi)} = \frac{k\pi - x}{k\pi(x-k\pi)} \prod_{n \neq k,0} \frac{n\pi - x}{n\pi} = -\frac{1}{k\pi} \prod_{n \neq k,0} \frac{n\pi - x}{n\pi}, $$ and taking the limit $x \to k \pi$ I have: $$ \tag 4 \lim_{x \to k \pi} \frac{\sin(x)}{x(x-k\pi)} = - \frac{1}{k\pi} \prod_{n\neq k,0} \frac{n-k}{n} = \frac{1}{k\pi}.$$ The last step comes from the following considerations: $$ \tag 5 \prod_{n \neq k,0} (n-k) = \frac{1}{(-k)} \prod_{n \neq 0} n, $$ $$ \tag 6 \prod_{n \neq k,0} n = \frac{1}{k} \prod_{n \neq 0} n,$$ $$ \tag 7 \prod_{n\neq k,0} \frac{n-k}{n} = \frac{k}{-k} = -1.$$

Where did I go wrong? To be more specific, I know how to obtain the result (2) directly, I'm trying to do it from the infinite product expansion as an exercise to try the formula.

If the problem is in my "splitting" up non-convergent components of the product, I would also like to know what is the correct handle this kind of calculation.

Answer 1

If you want to show it using Weierstrass product, this is a way$$\frac{\sin\left(x\right)}{x\left(x-k\pi\right)}=\frac{1}{x-k\pi}\underset{n=1}{\overset{\infty}{\prod}}\left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right)=\frac{1}{x-k\pi}\underset{n=1}{\overset{\infty}{\prod}}\left(\frac{\left(n\pi-x\right)\left(n\pi+x\right)}{n^{2}\pi^{2}}\right)=-\frac{k\pi+x}{k^{2}\pi^{2}}\underset{n=1,n\neq k}{\overset{\infty}{\prod}}\left(\frac{\left(n\pi-x\right)\left(n\pi+x\right)}{n^{2}\pi^{2}}\right)=-\frac{k\pi+x}{k^{2}\pi^{2}}\underset{n=1,n\neq k}{\overset{\infty}{\prod}}\left(1-\frac{x^{2}}{n^{2}\pi^{2}}\right)\underset{x\rightarrow k\pi}{\longrightarrow}-\frac{2}{k\pi}\underset{n=1,n\neq k}{\overset{\infty}{\prod}}\left(1-\frac{k^{2}}{n^{2}}\right).$$ It remain to show that $$\underset{n=1,n\neq k}{\overset{\infty}{\prod}}\left(1-\frac{k^{2}}{n^{2}}\right)=\frac{\left(-1\right)^{k-1}}{2}$$ and you can find a proof here How can I show that $\prod_{{n\geq1,\, n\neq k}} \left(1-\frac{k^{2}}{n^{2}}\right) = \frac{\left(-1\right)^{k-1}}{2}$? .

Answer 2

The products you are working with are not absolutely convergent : you can't sum the terms in the order you want. I'd guess that's why you don't get the correct answer.

Answer 3

It is better to use a translation: $$ \lim_{x\to k\pi}\frac{\sin(x)}{x(x-k\pi)}=\lim_{x\to 0}\frac{\sin(x+k\pi)}{x(x+k\pi)}=\frac{(-1)^k}{k\pi}\cdot\lim_{x\to 0}\frac{\sin x}{x}=\color{red}{\frac{(-1)^k}{k\pi}}.$$

Answer 4

You can't say that the two products in (5) are equal because neither one is convergent. You have the same issue with line (6).

In order to evaluate, $\prod_{n\ne k,0}\frac{n-k}n$, you might try to write it as $\prod_{n\ne k,0}\left(1-\frac kn\right)$.

Answer 5

The problem in my derivation is in taking separately the limits of divergent expressions, as already pointed out in the accepted answer. Here is a way to prove directly that $$ \prod_{n \neq 0,k} \frac{n-k}{n} = (-1)^{k+1}, \quad k \in \mathbb{Z}, $$ completing the last step of my derivation (the concept is very similar to this one):

$$ \prod_{n \neq 0,k} \frac{n-k}{n} = \lim_{N \to \infty} \frac {(-N-k)(-N-k+1)\cdots(-k-1)(-k+1)\cdots(-1)(+1)\cdots(N-k)} {-N(-N+1)\cdots(-1)(+1)\cdots(k-1)(k+1)\cdots N} \\ = \frac{(-1)^{N+k}}{-k} \lim_{N \to \infty} \left( \frac {(N+k)! (N-k)!} {-N(-N+1)\cdots(-1)(+1)\cdots(k-1)(k+1)\cdots N} \right) \\ = \frac{(-1)^{N+k}}{-k} k(-1)^N \lim_{N \to \infty} \frac {(N+k)! (N-k)!} {(N!)^2} = (-1)^{k+1} $$