In what sense is the "average" of $\sin(x)$ equal to $0$?

Question

I think it makes intuitive sense that if a phenomenon is described by simple sinusoidal oscillation, it is "on average" equal to its midrange, but I think I failed in trying to make that statement more rigorous.

Let's focus on $\sin(x)$. In a finite interval covering an integral number of cycles, it's obvious -- we can either throw a uniform distribution on the interval or just integrate and it'll be 0.

Unfortunately there's no (proper) infinite analog to the uniform distribution, and even if we use the improper version (density = 1 everywhere), the infinite integral of $\sin(x)$ strikes me as divergent because its partial sums diverge. One workaround to this I came up with is:

$$\intop_{-\infty}^{\infty}\sin(x)dx = \sum_{k\in\mathbb{N}}\left(\intop_{2\pi k}^{2\pi(k+1)}\sin(x) dx\right)=0$$

But that strikes me as a bit fishy (e.g., we should require that the LHS exists before throwing an equals sign down, correct?).

Is there any sense, then, in which the average of $\sin(x)=0$?

Take a pendulum. I conjecture that if asked to guess where the pendulum is at a random time, under any symmetric, $0$-modal loss function, your best bet would be to choose that it's in the middle.

Answer 1

We can define an average for an unbounded domain like this :

If for any sequence of intervals $I_1 \subset I_2 \subset \cdots $ that "converge" to $\Bbb R$ (i.e, $\bigcup_{n\in \Bbb N} I_n = \Bbb R$), you have $$\lim_{n \to + \infty} \frac{1}{|I_n|} \int_{I_n} \sin(x) dx = 0$$

then we say that the average of $\sin(x)$ over $\Bbb R$ is $0$

And you can notice that it's compatible with the usual definition of an average for bounded intervals $I$

Answer 2

Here is one simple way to make it rigorous: if $f(x)=a\sin(bx+c)$ is a sine wave, then we have

$$\lim_{t\to\infty}\frac{1}{t}\int_0^tf(x)dx = 0$$