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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!

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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.

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  • $\begingroup$ Ah, I see. It's more complicated than I thought it is actually, with the speed of the convergence taken into account! If I could trouble you for one more thing; I'm not sure how did you come to the result of the integral to be $$M=\lim\limits_{T \to \infty} \dfrac{1}{2T}\int_{-T}^{T} \text{dt} f(t) = \frac{f(+\infty) + f(-\infty)}{2},$$ maybe you can give me some hint? Thanks in any case! $\endgroup$ – Lajka Mar 8 '14 at 13:16
  • $\begingroup$ The idea is that nearly all of the real line is "near" $+\infty$ and $-\infty$, and everything else is negligible. For every $\epsilon > 0$, there is an $N$ such that $t > N$ implies $|f(x) - f(+\infty)| < \epsilon$. The positive half of the integral breaks apart into $$\int_0^N f(t) \, dt + \int_N^T f(t) \, dt$$ The first part half is constant, and you can use the mean value theorem on the second part. It may help to rewrite the integral as $$\int_0^N f(t) - f(\infty) \, dt + \int_N^T f(t) - f(\infty) \, dt + \int_0^T f(\infty) \, dt$$ $\endgroup$ – user14972 Mar 8 '14 at 16:50
  • $\begingroup$ I am sorry, you just confused me even more. :D Maybe I should've mentioned I'm just a poor electrical engineer and not a math student. The only thing I see here is that, should your equation be true, this has to happen $$M=\dfrac{1}{2T}\int_{-T}^{T} \text{dt} f(t) = \frac{f(+T) + f(-T)}{2},$$ and then, when we let T to go to infinity, we get your equation. I just can't understand how this first equation (without the limit) came to be. $\endgroup$ – Lajka Mar 9 '14 at 23:39
  • $\begingroup$ The only thing I can think about here, is utilizing the N-L theorem $$\dfrac{1}{2T} \int_{-T}^{T} \text{dt} f(t) = \frac{F(+T) - F(-T)}{2T},$$ (F is antiderivative), which doesn't really tell me much, nor gets me closer to your equation. Maybe I could trouble you for some more tips? $\endgroup$ – Lajka Mar 9 '14 at 23:46
  • $\begingroup$ @Lajka: A common trick in analysis is to split a problem apart, to express it's dominant features most directly. My rewrite of the integral $\int_0^T f(t) \, dt$ above breaks it into three parts: the third part has an actual constant integrand, the second part has an integrand we know is bounded by $\epsilon$, and the first part is what is left over, but happens to be constant with respect to $T$. We can do a similar thing for the negative half of the integral. Thus, when computing the limit of $\frac{1}{2} \int_{-T}^T f(t) \, dt$, we rewrite the integral in the way I describe, and can... $\endgroup$ – user14972 Mar 9 '14 at 23:48
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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 $c$.

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Yet another Counter example, this time non constant

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

Quite obviously it's mean should be $c$

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  • $\begingroup$ Thanks for that! Makes me now wonder what would happen if the boundaries of the integral would converge with different speeds... $\endgroup$ – Lajka Mar 8 '14 at 13:16

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