reference parseval theorem limited signal
I don't think you will find a reference for this. Your are simply stating that "integrating over a whole bunch of zeros results in 0". In other words, if 
$$x(t) = 0, x\not\in [0,T]$$
then
$$ \int_{-\infty}^{+\infty}x(t) dt = \int_{-\infty}^{0}x(t) dt + \int_{0}^{T}x(t) dt + \int_{T}^{+\infty}x(t)dt = \int_{0}^{T}x(t) dt $$
since
$$ \int_{-\infty}^{0}x(t) dt = \int_{-\infty}^{0}0 dt=0$$ and

$$\int_{T}^{+\infty}x(t)dt = \int_{T}^{+\infty}0dt = 0$$

If that doesn't work, than I don't know what would