The energy theorem
The energy in a wave is proportional to the square of its amplitude. For a wave of complex shape, the energy in one period will be proportional to $\int_0^Tf^2(t)\,dt$. We can also relate this energy to the Fourier coefficients. We write  \begin{equation} \label{Eq:I:50:22} \int_0^Tf^2(t)\,dt = \int_0^T\biggl[a_0 + \sum_{n = 1}^\infty a_n\cos n\omega t + \sum_{n = 1}^\infty b_n\sin n\omega t\biggr]^2\,dt. \end{equation} 
 \begin{gather} \label{Eq:I:50:22} \int_0^Tf^2(t)\,dt =\\ \int_0^T\biggl[a_0 + \sum_{n = 1}^\infty a_n\cos n\omega t + \sum_{n = 1}^\infty b_n\sin n\omega t\biggr]^2\,dt.\notag \end{gather}  When we expand the square of the bracketed term we will get all possible cross terms, such as $a_5\cos5\omega t\cdot a_7\cos7\omega t$ and $a_5\cos5\omega t\cdot b_7\sin7\omega t$.  We have shown above, however, [Eqs. (50.11) and (50.12)] that the integrals of all such terms over one period is zero.  We have left only the square terms like $a_5^2\cos^2 5\omega t$. The integral of any cosine squared or sine squared over one period is equal to $T/2$, so we get  \begin{align} \int_0^Tf^2(t)\,dt &= Ta_0^2 + \frac{T}{2}\, (a_1^2 + a_2^2 + \dotsb + b_1^2 + b_2^2 + \dotsb)\notag\\[.5ex] \label{Eq:I:50:23} &= Ta_0^2 + \frac{T}{2}\sum_{n = 1}^\infty(a_n^2 + b_n^2). \end{align} 
 \begin{gather} \label{Eq:I:50:23} \int_0^Tf^2(t)\,dt =\\[.5ex] Ta_0^2 + \frac{T}{2}\,(a_1^2 + a_2^2 + \dotsb + b_1^2 + b_2^2 + \dotsb)= \notag\\[.5ex] Ta_0^2 + \frac{T}{2}\sum_{n = 1}^\infty(a_n^2 + b_n^2).\notag \end{gather}   This equation is called the “energy theorem,” and says that the total energy in a wave is just the sum of the energies in all of the Fourier components. For example, applying this theorem to the series (50.19), since $[f(t)]^2 = 1$ we get \begin{equation*} T = \frac{T}{2}\cdot\biggl(\frac{4}{\pi}\biggr)^2\biggl( 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dotsb\biggr), \end{equation*} so we learn that the sum of the squares of the reciprocals of the odd integers is $\pi^2/8$. In a similar way, by first obtaining the Fourier series for the function $f(t)=(t-T/2)^2$ and using the energy theorem, we can prove that $1 + 1/2^4 + 1/3^4 + \dotsb$ is $\pi^4/90$, a result we needed in Chapter 45.