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

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 $$T \to \infty$$ ?

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.

• 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? Apr 5 '20 at 18:44
• I did not think of that, let me try. Apr 5 '20 at 18:44

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 $$f(x)=x$$, $$\lim_{T\to\infty}\frac{1}{T}\int_0^T f(s)ds=\lim_{T\to\infty}\frac{T^2}{2T}=\infty$$

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.

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

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

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