Mean value over an infinite interval

Question

I just need a sanity check, been thinking about this all morning.

If we use the Mean Value Theorem on a function over the infinite interval (suppose the function's domain is unbounded), i.e.

$$M=\lim\limits_{T \to \infty} \dfrac{1}{2T}\int_{-T}^{T} \text{dt} f(t)$$

There is no way that M can be finite right? My intuition tells me it's either zero or infinite, but I wanted another opinion; oddly enough, I wasn't able to google it.

Thanks!

Answer 1

Trying special cases is always useful. $f(t) = 1$ is one of the simplest options. And there's a rather intuitive idea of what its mean should be too!

If $f$ has a limit at $+\infty$ and $-\infty$, then I'm pretty sure the mean turns out to be $$\frac{f(+\infty) + f(-\infty)}{2}$$ whenever defined. (i.e. this formula makes no prediction about the mean of $f(x) = x$ or $f(x) = x+1$)

Incidentally, consider allowing both endpoints to vary independently:

$$ \lim_{(s,t) \to (+\infty, +\infty)} \frac{1}{s+t} \int_{-s}^t f(x)\,dx$$

Your definition resembles a sort of "Cauchy principal value" of this one.

Answer 2

Take $f(t)$ to be a constant, say $c$. Then the integral equals $c$ for all $T$, thus the limit and $M$ are equal to $c$.

Answer 3

Yet another Counter example, this time non constant

$$ f(x) = atan(x) + c $$

Quite obviously it's mean should be $c$