Calculation of the power of a signal Suppose we want to calculate the power of the signal $y(t) = m(t)\cos(\omega_c t)$, where $m(t)$ has zero mean, and the power of $m(t)$ is P watts.
It is easy to show that the power of $y(t)$ is $P/2$ by first calculating the power spectral density (PSD) of $y(t)$ and then integrating it from $-\infty$ to $\infty$. However, if I go to the original definition of the power of a signal, I get stuck as follows:
$$
\begin{align}
P_y &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} y^2(t)dt\\
&=\lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} m^2(t)\cos^2(\omega_c t)dt\\
&=\lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} m^2(t)\left[\frac{1+\cos(2 \omega_c t)}{2}\right]dt\\
&=\frac{P}{2} + \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} \frac{m^2(t)\cos(2 \omega_c t)}{2}dt\\
\end{align}
$$
I am unable to find any reason for the integral in the last line to be zero. Is there any reason that the integral has to be zero?
 A: Note that
$$\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2}m^2(t)\cos(2\omega_c t)dt$$
is the DC value of the function $m^2(t)\cos(2\omega_c t)$. Assuming that $m(t)$ is a band limited low pass signal with a band edge $\omega_0<\omega_c$ (i.e., $m^2(t)$ has a band edge $2\omega_0<2\omega_c$), then the above DC term equals zero.
If $m(t)$ is not ideally band limited, the DC term of the modulated signal is very likely to be negligible, but the integral will not evaluate to (exactly) zero.
You can also determine the power spectrum of $y(t)$ to arrive at the same result. Let $M(\omega)$ be the power spectrum of $m(t)$:
$$M(\omega)=\lim_{T\rightarrow\infty}\frac{1}{T}\left|\int_{-T/2}^{T/2}m(t)e^{-j\omega t}dt\right|^2$$
The power spectrum of $y(t)$ is
$$Y(\omega)=\lim_{T\rightarrow\infty}\frac{1}{T}\left|\int_{-T/2}^{T/2}y(t)e^{-j\omega t}dt\right|^2=
\lim_{T\rightarrow\infty}\frac{1}{T}\left|\int_{-T/2}^{T/2}m(t)\cos(\omega_c t)e^{-j\omega t}dt\right|^2=
\lim_{T\rightarrow\infty}\frac{1}{T}\left|\int_{-T/2}^{T/2}m(t)\frac12 (e^{j\omega_ct}+e^{-j\omega_ct})e^{-j\omega t}dt\right|^2=
\lim_{T\rightarrow\infty}\frac{1}{T}\frac14
\left|\int_{-T/2}^{T/2}m(t)e^{j\omega_ct}e^{-j\omega t}dt+
\int_{-T/2}^{T/2}m(t)e^{-j\omega_ct}e^{-j\omega t}dt\right|^2=
\lim_{T\rightarrow\infty}\frac{1}{T}\frac14\left[
\left|\int_{-T/2}^{T/2}m(t)e^{j\omega_ct}e^{-j\omega t}dt\right|^2+
\left|\int_{-T/2}^{T/2}m(t)e^{-j\omega_ct}e^{-j\omega t}dt\right|^2+
2\Re\left\{\int_{-T/2}^{T/2}m(t)e^{j\omega_ct}e^{-j\omega t}dt\cdot
\int_{-T/2}^{T/2}m^*(t)e^{j\omega_ct}e^{j\omega t}dt\right\}
\right]=\\
\frac14\left[M(\omega-\omega_c)+M(\omega+\omega_c)\right]+
\frac12\lim_{T\rightarrow\infty}\frac{1}{T}\Re\left\{\ldots\right\}$$
The power of $y(t)$ is obtained by integrating the power spectrum. Integrating the first term $\frac14\left[M(\omega-\omega_c)+M(\omega+\omega_c)\right]$ gives $P/2$, where $P$ is the power of $m(t)$. But there is also the last term which is generally non-zero. However, for almost all practical purposes this term can be neglected, because, as shown above, it depends on the spectral content of $m(t)$ at the modulation frequency $\omega_c$, which is usually either zero or negligibly small.
A: You chose the wrong bounds of integration. Rather than integrating across $[-\frac{T}{2},\frac{T}{2}]$, integrate across $[0,T]$. This will give a non-zero value. The choice of bounds shouldn't matter, but it does because of the way the integral will account for the negative component of the signal.
