Find a g such that $\lim_{T\to\infty}\frac{1}{T}\int_{0}^{T} g(\tau)d\tau$ does not exist

Question

I have to solve the following: find a continuous and bounded function $g:[0,\infty) \to \mathbb{C}$ s.t. the limit $$\lim_{T\to\infty}\frac{1}{T}\int_{0}^{T} g(\tau)d\tau$$ does not exist.

At the beginning I was thinking of looking for a function such as $$g(\tau) = sin\tau + \tau cos\tau$$

so that $\int_{0}^{T} g(\tau)d\tau = Tsin(T)$ and its limit to infinite does not exist due to the oscillating behavior of the sine, but then I realized such a $g(\tau)$ is not bounded.

So what should I look for? It seems that I cannot ''cancel'' the term $\frac{1}{T}$ using a bounded function, so which should be the right way of approaching the problem?

Answer 1

Let $g$ be something like $g(x)=\sin( \log(1+x))$.