I am studying signals and system.
I learned that \begin{align} P&=\lim_{L\to\infty} \frac 1{2L} \int_{-L}^{L} |x(t)|^2 dt\\ P&=\frac 1{T} \int_{<T>} |x(t)|^2 dt ~~~\mbox{, P could be solved with this easier equation when } x(t) \mbox{ is periodic.} \end{align}
I made a artificial signal myself like the following. \begin{align} x(t)&=t &x&\in(0, T)\\ x(t+T)&=x(t) &x&\in(-\infty, \infty) \end{align}
using the definition of $P$,
\begin{align} P &= \lim_{L\to\infty} \frac{1}{2L} \left(\int_{-L}^0 |t+T|^2 dt + \int_0^L |t|^2 dt \right)\\ &= \lim_{L\to\infty} \frac{1}{2L} \left(\left[\frac{(t+T)^3}{3}\right]^0_{-L} + \left[ \frac{t^3}{3} \right]_0^L\right)\\ &= \lim_{L\to\infty} \frac{1}{2L} \left(\frac{T^3 - (T-L)^3}{3}+\frac{L^3}3\right)\\ &= \lim_{L\to\infty} \frac{1}{2L} \cdot \frac{3T^2L-3TL^2+2L^3}{3}\\ &= \lim_{L\to\infty} \left( \frac{T^2}{2}-\frac{TL}{2}+\frac{L^2}{3} \right)=\infty \end{align}
using the modified equation because $x(t)$ is periodic,
\begin{align} P &= \frac{1}{T} \int_0^T |t|^2 dt \\ &= \frac{1}{T} \left[\frac{t^3}{3}\right]^T_0\\ &= \frac{T^2}{3} \end{align}
I felt so weird. What's wrong with my process. Thank you for reading my question.