If $ E_\infty $ is finite, then $ P_\infty $ is always zero
$ P_\infty\equiv\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T|x(t)|^2dt $
$ P_\infty\equiv(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}\int_{-T}^T|x(t)|^2dt) $
Because $ E_\infty\equiv\lim_{T\to\infty}\int_{-T}^T|x(t)|^2dt $, it follows that by substitution
$ P_\infty=(\lim_{T\to\infty}\frac{1}{2T})*E_\infty $
$ P_\infty=(\lim_{T\to\infty}\frac{1}{2T})*(\lim_{T\to\infty}E_\infty) $
$ P_\infty=\lim_{T\to\infty}\frac{E_\infty}{2T} $
This limit will always evaluate to zero as long as $ E_\infty $ is finite.
$ \therefore $ If $ E_\infty $ is finite, then $ P_\infty $ is always zero $ \square $
-Adam Siembida
