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Original

Does the following exist: $$\lim_{T \to \infty} \frac{1}{T} \int_0^T f(s) ds$$ I know the answer if the limit was $T \to 0$: the limit then becomes the derivative of the integral which by the Fundamental Theorem of Calculus is $f(T)$. But can we talk about the limit when $T \to \infty$ ?

Edit and Answer

Since the question was closed and I had a more thorough thinking about it, I will rephrase the question like this:

Under what conditions for $f$ does $\lim_{T \to \infty} \frac{1}{T} \int_0^T f(s) ds$ exist?

I think most useful and easy to compute sufficient condition for the limit above to exists can be obtained if we use L'Hôpital's rule:

$$\lim_{T \to \infty} \frac{\int_0^T f(s) ds}{T} = \lim_{T \to \infty} \frac{\frac{d}{dT}\int_0^T f(s)ds}{\frac{d}{dT}T} = \lim_{T \to \infty} \frac{f(T)}{1} $$

i.e. the limit exists if $\lim_{T \to \infty} f(T)$ exists.

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    $\begingroup$ Have you tried simple choices for $f$, like $f(s) = T$, $f(s) = s$, $f(s) = \mathrm{e}^s$, and perhaps others, to get some intuition for what is happening? $\endgroup$ Apr 5 '20 at 18:44
  • $\begingroup$ I did not think of that, let me try. $\endgroup$
    – baibo
    Apr 5 '20 at 18:44
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It depends on $f$. For example, for $f(x)=1$, $$\lim_{T\to\infty}\frac{1}{T}\int_0^T f(s)ds=\lim_{T\to\infty}\frac{T}{T}=1$$ but for $f(x)=x$, $$\lim_{T\to\infty}\frac{1}{T}\int_0^T f(s)ds=\lim_{T\to\infty}\frac{T^2}{2T}=\infty$$

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If you don't pose any restrictions on $f$, then the limit does not necessarily exist. Consider for example $f(s) := e^s$, then we have

$$ \lim_{T \rightarrow \infty} \frac{1}{T} \int^T_0 e^s ds = \lim_{T \rightarrow \infty} \frac{e^T - 1}{T} \rightarrow \infty ~~.$$

How did I come up with this? The expression $\frac{1}{T} \int^T_0 e^s ds$ can be thought of as the average of the function $f$ on the interval $[0,T]$. Now take some function which gets very large for $s \rightarrow \infty$ and you have your candidate.

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Let's assume $$\int_0^Tf(s)ds=F(T)-F(0)$$

Then $$\lim_{T\to\infty}\frac1T\int_0^Tf(s)ds=\lim_{T\to\infty}\frac{F(T)-F(0)}{T}=\lim_{T\to\infty}\frac{F(T)}{T} \text{(why?)}$$

For this limit to exist generally, every differentiable function $F$ must have $\lim_{T\to\infty}\frac{F(T)}{T}=L$ for some $L$ (can be infinite). Testing a few functions shows this is not the case

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Hint: think of $f(x)=x\sin(x)$. Does the limit exist for this function?

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  • $\begingroup$ No, not in this case. I see now that I should have tried a few simple cases to get that it depends on $f$. $\endgroup$
    – baibo
    Apr 5 '20 at 18:50
  • $\begingroup$ @baibo Yep, under some restrictions, maybe the limit exists, but one would have to think about those carefuly! $\endgroup$ Apr 5 '20 at 18:53
  • $\begingroup$ @baibo Why the downvote? I illustrated an example that answers your question $\endgroup$ Apr 5 '20 at 21:37
  • $\begingroup$ The downvote wasn't from me $\endgroup$
    – baibo
    Apr 6 '20 at 8:16

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