Defining an average in time for a density in time I'm reading the text book "Time-Frequency Analysis" by Leon Cohen and I've made my way through a decent portion. There is however a conceptual issue I keep coming back to. The book states:


  If we consider $\lvert s(t)\rvert^2$ as a density in time, the average
  time can be defined in the usual way any average is defined:
  
  $$  \langle t \rangle  = \int t \lvert s(t)\rvert^2dt$$


I don't understand conceptually why multiplying by $t$ in the integration will provide us with an average. I didn't know that was the typical way to define an average either despite the books insistence that this is common knowledge (which it very may well be I'm just not interpreting it correctly). I've always thought about an average as dividing by some base, say by the total period of time $T$ NOT multiplying by $t$ inside the integral. It states earlier that:


  $\lvert s(t)\rvert^2$ is the energy per unit time.


So I was thinking that perhaps mathematically multiplying by $t$ which has units of $s$ would get rid of the per unit time portion, but the units of $\lvert s(t)\rvert^2$ aren't actually $v^2/s$ as that would be the derivative, it's just $v^2$ so that doesn't make much sense (assuming the signal is an electrical signal and the unit time is seconds). Can someone point me in the right direction on how to think about this? I know it's a simple concept but I'm just having trouble making it over this hurdle. Thanks!
I think it is implied that $\int_{-\infty}^{\infty}|s(t)|^2 dt = 1$ (which is why the author refers to $|s(t)|^2$ as a density).

I agree with your sentiment "I've always thought about an average as dividing by some base", but in this case, the base would be 1, so no division is needed. If you were to plot $|s(t)|^2$ as a function of $t$, and think of this plot as a histogram, the the integral:

$$\int_{-\infty}^{\infty}t|s(t)|^2 dt $$
will tell you the average value, since each value of $t$ is weighted by the density function.