Originally Posted by

**chisigma** The limit...

$\displaystyle \lim_{t \rightarrow \infty} \int_{0}^{t} \sin \tau\cdot d\tau$ (1)

... doesn't exist but exists the limit...

$\displaystyle \lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} \sin \tau\cdot d\tau = \langle \sin t \rangle = 0$ (2)

Welll...yes, in a rather trivial way: since INT{0,t} sin(s) ds = -cos(t) + 1,

then lim (1/t)[1 - cos(t)] = 0 when t --> oo since 1 - cos t is bounded and (1/t) --> 0 .

This though doesn't seem to appropiately address the OP's question.

Tonio

In general if $\displaystyle f(*)$ is the sum of a finite number of periodic functions *not necessarly with the same period* the limit...

$\displaystyle \lim_{t \rightarrow \infty} \frac{1}{t}\cdot \int_{0}^{t} f(\tau)\cdot d\tau = \langle f(t) \rangle $ (3)

... exists and $\displaystyle \langle f(*) \rangle$ is called [improperly] 'mean value of $\displaystyle f(*)$'. It is...

$\displaystyle \langle f(t) \rangle = \sum_{n} \langle f_{n}(t) \rangle$ (4)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$