Average Power for "Infinite period of time" of a periodic signal
The bounds are not actually important mathematically, but they help with meaning.

You could have written \$P=\lim_{\tau\to\infty}\frac{1}{2\tau}\int_{-\tau}^\tau p(t)dt\$ and gotten the same results.  However, if you consider all of the different power functions:


Finite bound: \$P=\frac{1}{\tau}\int_{T}^{T+\tau} p(t)dt\$
One sided infinite: \$P=\lim_{\tau\to\infty}\frac{1}{\tau}\int_{0}^\tau p(t)dt\$
Two sided infinite: \$P=\lim_{\tau\to\infty}\frac{1}{\tau}\int_{\frac{-\tau}{2}}^\frac{\tau}{2} p(t)dt\$


We see that \$\tau\$ has the same meaning of "period" or size of integration window in all of these cases.  This is convenient for looking for parallels between the different equations.  If we integrated the two sided infinite equation from \$-\tau\$ to \$\tau\$, it would obscure the connection between these different forms.

You'll also see this happen later with many transforms, like the Fourier transform.  We could change the way we define things to avoid some annoying \$2\pi\$ terms that arise, but that would then obscure the connection between the Fourier transform of some data and its original form.