# In what sense is the “average” of $sin(x)$ equal to 0?

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:

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

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.

• Are you looking for an answer that doesn't invoke any Fourier analysis? – Ryan Dec 24 '15 at 20:37
• @Ryan I don't have any prior about the form of the answer – MichaelChirico Dec 24 '15 at 20:38

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 $I$

• @Travis Or you could think of it as the Moore-Smith limit of a net defined on the directed set consisting of real intervals ordered by inclusion. – bof Dec 24 '15 at 21:45
• @Travis : Yes, I agree, I will modify the wording. – Tryss Dec 24 '15 at 23:47
• This doesn't fully capture bounded oscillatory nature of $\sin x$. Your approach would assign average value of $0$ to $x^{1/2}\sin(x)$ (and $\frac 1 {1+|x|}\sin x$) while the three cases are qualitatively different. I'n not sure as to what other simple metric (apart from $\sup$ norm) can capture these differences – A.S. Dec 25 '15 at 0:36
• Equivalently, taking a limit of uniform distributions, the limit of which is degenerate. – MichaelChirico Dec 25 '15 at 2:28
• @A.S : it's hard to have a single notion that capture all the differences. Here I'm not sure the bounded oscillatory nature is what was important but more "how can we assign an average on an unbounded domain". I give here a possible answer : if, no matter what sequence of nested invervals you choose that "converge" to the whole domain, the value is the same, then this is defined as the average on the whole domain. (yes, I didn't exactly write this, so I'll slightly modify my answer) – Tryss Dec 25 '15 at 3:47

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

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