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