149.

Let us now calculate the total energy which is absorbed by the oscillator in the time from $t=0$ to $t=\tau$, where ${\omega }_0\tau \text{ is large}.\mathrm{\backslash Label}[eqn](239)\text{upshape (239)}$ According to equation [eqn:(234)] (234), it is given by the integral ${\int }_0^{\tau }{⋿}_z\frac{df}{dt}dt,\mathrm{\backslash Label}[eqn](240)\text{upshape (240)}$ the value of which may be obtained from the known expression for ${⋿}_z$ [eqn:(235)] (235) and from $\frac{df}{dt}={\sum }_1^{\infty }[a_n(-\omega \sin \omega t+{\omega }_0\sin {\omega }_{0}t)+b_n(\omega \cos \omega t-\omega \cos {\omega }_{0}t)].\mathrm{\backslash Label}[eqn](241)\text{upshape (241)}$ By multiplying out, substituting for $a_n$ and $b_n$ their values from [eqn:(238)] (238), and leaving off all terms resulting from the multiplication of two constants $A_n$ and $B_n$, this gives for the absorbed energy the following value:

$$\begin{gathered} \frac{1}{L}{\int }_0^{\tau }dt{\sum }_1^{\infty }[\frac{A_n^2}{{\omega }_0^2-{\omega }^2}\cos \omega t(-\omega \sin \omega t+{\omega }_0\sin {\omega }_{0}t) \\ +\frac{B_n^2}{{\omega }_0^2-{\omega }^2}\sin \omega t(\omega \cos \omega t-\omega \cos {\omega }_{0}t)].\mathrm{\backslash Label}[eqn](241a)\text{upshape (241a)} \end{gathered}$$

In this expression the integration with respect to $t$ may be performed term by term. Substituting the limits $\tau$ and $0$ it gives

$\&\frac{1}{L}{\sum }_1^{\infty }\frac{A_n^2}{{\omega }_0^2-{\omega }^2}[-\frac{{\sin }^2\omega \tau }{2}+{\omega }_0(\frac{{\sin }^2{\displaystyle \frac{{\omega }_0+\omega }{2}}\tau }{{\omega }_0+\omega }+\frac{{\sin }^2{\displaystyle \frac{{\omega }_0-\omega }{2}}\tau }{{\omega }_0-\omega })]$

$+\&\frac{1}{L}{\sum }_1^{\infty }\frac{B_n^2}{{\omega }_0^2-{\omega }^2}[\phantom{-}\frac{{\sin }^2\omega \tau }{2}-\omega (\frac{{\sin }^2{\displaystyle \frac{{\omega }_0+\omega }{2}}\tau }{{\omega }_0+\omega }-\frac{{\sin }^2{\displaystyle \frac{{\omega }_0-\omega }{2}}\tau }{{\omega }_0-\omega })].$

In order to separate the terms of different order of magnitude, this expression is to be transformed in such a way that the difference ${\omega }_0-\omega$ will appear in all terms of the sum. This gives

$$\begin{gathered} \frac{1}{L}{\sum }_1^{\infty }\frac{A_n^2}{{\omega }_0^2-{\omega }^2}[\frac{{\omega }_0-\omega }{2({\omega }_0+\omega )}{\sin }^2\omega \tau +\frac{{\omega }_0}{{\omega }_0+\omega }\sin \frac{{\omega }_0-\omega }{2}\tau ·\sin \frac{{\omega }_0+3\omega }{2}\tau \\ +\frac{{\omega }_0}{{\omega }_0-\omega }{\sin }^2\frac{{\omega }_0-\omega }{2}\tau ]\phantom{.} \end{gathered}$$

$$\begin{gathered} +\frac{1}{L}{\sum }_1^{\infty }\frac{B_n^2}{{\omega }_0^2-{\omega }^2}[\frac{{\omega }_0-\omega }{2({\omega }_0+\omega )}{\sin }^2\omega \tau -\frac{\omega }{{\omega }_0+\omega }\sin \frac{{\omega }_0-\omega }{2}\tau ·\sin \frac{{\omega }_0+3\omega }{2}\tau \\ +\frac{\omega }{{\omega }_0-\omega }{\sin }^2\frac{{\omega }_0-\omega }{2}\tau ]. \end{gathered}$$

The summation with respect to the ordinal numbers $n$ of the Fourier's series may now be performed. Since the fundamental period $𝖳$ of the series is extremely large, there corresponds to the difference of two consecutive ordinal numbers, $\Delta n=1$ only a very small difference of the corresponding values of $\omega$, $d\omega$, namely, according to [eqn:(236)] (236), $\Delta n=1=𝖳d\nu =\frac{𝖳d\omega }{2\pi },\mathrm{\backslash Label}[eqn](242)\text{upshape (242)}$ and the summation with respect to $n$ becomes an integration with respect to $\omega$.

The last summation with respect to $A_n$ may be rearranged as the sum of three series, whose orders of magnitude we shall first compare. So long as only the order is under discussion we may disregard the variability of the $A_n^2$ and need only compare the three integrals

$$\begin{gathered} {\int }_0^{\infty }d\omega \frac{{\sin }^2\omega \tau }{2({\omega }_0+\omega {)}^2}=J_1, \\ {\int }_0^{\infty }d\omega \frac{{\omega }_0}{({\omega }_0+\omega {)}^2({\omega }_0-\omega )}\sin \frac{{\omega }_0-\omega }{2}\tau ·\sin \frac{{\omega }_0+3\omega }{2}\tau =J_2, \\ \mathrm{\backslash intertext}and{\int }_0^{\infty }d\omega \frac{{\omega }_0}{({\omega }_0+\omega )({\omega }_0-\omega {)}^2}{\sin }^2\frac{{\omega }_0-\omega }{2}\tau =J_3. \end{gathered}$$