Prove that $\sin x$ is not of bounded variation, $\sin x\not\in BV$

Question

I already know that $\sin x\in BV([a,b])$ for $-\infty<a<b<\infty$.

However, I have failed to see why $\sin x\notin BV$. Please help.

Answer 1

Hint. Let $x_n = \frac{\pi(2n + 1)}{2}$ for $n\in\mathbb{N}$. Then $\{x_n\}$ is a strictly increasing sequence and $\sin(x_n)=(-1)^n$. Now consider the sum $$\sum_{n=1}^N \left| \sin(x_n) - \sin(x_{n-1})\right|=\sum_{n=1}^N 2=2N.$$ What may we conclude? Recall the definition of bounded variation function and take a look at Functions of bounded variation on $\Bbb R$

Answer 2

The derivative of $sin(x)$ is $cos(x)$. The arc length of its derivative $ \int_a^b |cos(x)| dx$ can thus be made arbitrarily large as we consider increasingly large closed intervals $[-n \pi, n \pi]$