Can the total energy of a signal diverge while its average power converges to zero?

Question

If the integral used to calculate the total energy for a continuous time real signal converges, its average power is equal to zero. But could the average power still equal zero if the total energy diverges?

Equivalently, could this limit evaluate to zero

$$P_\infty=\lim_{T \to \infty}\frac{1}{2T} \int_{-T}^{T}\left| x(t) \right|^2dt=0$$

if this limit diverges

$$E_\infty=\lim_{T \to \infty} \int_{-T}^{T}\left| x(t) \right|^2dt=\infty$$

Answer 1

That's indeed possible, at least in theory. Just come up with a signal which when squared and integrated diverges as $t\to\infty$, but slower than linearly. E.g.,

$$x(t)=\frac{1}{\sqrt[4]{|t|}}$$

$$\int_{-T}^T|x(t)|^2dt=\int_{-T}^T\frac{dt}{\sqrt{|t|}}=4\sqrt{T}$$

Consequently, $E_x\to\infty$ but

$$P_x=\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T\frac{dt}{\sqrt{|t|}}=\lim_{T\to\infty}\frac{2}{\sqrt{T}}=0$$