Does the improper integral exist?

Question

I need to find a continuous and bounded function $\mathrm{f}(x)$ such that the limit $$ \lim_{T\to\infty} \frac{1}{T}\, \int_0^T \mathrm{f}(x)~\mathrm{d}x$$ doesn't exist.

I thought about $\mathrm{f}(x) = \sin x$ but I am not sure if the fact that we divide by $T$ may some how make it converge to zero.

What do you think ?

Answer 1

$ \sin(x) $ won't do, but it's only a tad trickier. $$ f(x) = \sin\left(\ln(x)\right) $$ should do the job (you can integrate it exactly by elementary techniques and then show it is $ \mathcal{O}(T) $).

Note that the function is bounded and continuous in $ (0,+\infty) $.

Answer 2

The integral has to diverge to infinity if this limit is to not exist, (of course any other limit will give zero overall). Given the type $``\dfrac{\infty}{\infty}"$ limit, if we apply L'Hopitals; $$\displaystyle\lim_{T\to \infty}\dfrac{1}{T}\displaystyle\int_{0}^{T}f(x)\ dx = \lim_{T\to \infty} f(T)$$

This makes it clear that the limit doesn't exist iff $f$ is not bounded, so there is no suitable function.

Answer 3

The idea here is simple, but the details laborious. Keep the function at $+1$ long enough so the average hits some positive value, then keep the function at $-1$ long enough so the average hits some negative value and repeat.

Let $t_n = 2^{n+2}-4$. Let $f_n(t) = \begin{cases} t-t_n, & t \in [t_n,t_n+1] \\ 1 , & t \in (t_n+1,t_{n+1}-1) \\ 1-(t-(t_{n+1}-1)) , & t \in [t_{n+1}+1, t_n] \\ 0, \text{otherwise}\end{cases}$.

Note that $f_n$ has support $(t_n,t_{n+1})$, and $\int_0^{t_{n+1}} f_n = t_n+3$.

Let $\phi(t) = \sum_{k=0}^\infty (-1)^kf_k(t)$. Then \begin{eqnarray} \int_0^{t_{n+1}} \phi_n &=& \sum_{k=0}^n (-1)^k (t_k+3) \\ &=& \sum_{k=0}^n (-1)^k (2^{k+2}-1) \\ &=& \sum_{k=0}^n (-2)^{k+2}- \sum_{k=0}^n (-1)^k \\ &=& {4 \over 3} (1-(-2)^{n+1})- {1 \over 2} (1 - (-1)^{n+1}) \end{eqnarray} Then \begin{eqnarray} \sigma_n &=& {1 \over t_{n+1} }\int_0^{t_{n+1}} \phi_n \\ &=& {4 \over 3} { (1-(-2)^{n+1})- {1 \over 2} (1 - (-1)^{n+1}) \over 2^{n+3}-4} \\ &=& {1 \over 3}{ (1-(-2)^{n+1})- {1 \over 2} (1 - (-1)^{n+1}) \over 2^{n+1}-1} \\ &=& {1 \over 3} { {1 \over 2^{n+1} } - (-1)^{n+1} - {1 \over 2^{n+2} (1 - (-1)^{n+1} }\over 1 - {1 \over 2^{n+1} } } \end{eqnarray} We see that $\liminf_n \sigma_n = - {1 \over 3}$, $\limsup_n \sigma_n = {1 \over 3}$, hence the limit does not exist.

Answer 4

What about $\mathrm{f}(x)=T$?

$$ \lim_{T\to\infty} \frac{1}{T}\, \int_0^T \mathrm{f}(x)~\mathrm{d}x=\lim_{T\to\infty} \frac{1}{T}\, \int_0^T T~\mathrm{d}x=\lim_{T\to\infty} \frac{1}{T}\, T^2=\infty$$