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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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