149.

The evaluation of these integrals is greatly simplified by the fact that, according to [eqn:(239)] (239), ${\omega }_0\tau$ and therefore also $\omega \tau$ are large numbers, at least for all values of $\omega$ which have to be considered. Hence it is possible to replace the expression ${\sin }^2\omega \tau$ in the integral $J_1$ by its mean value $\frac{1}{2}$ and thus we obtain: $J_1=\frac{1}{4{\omega }_0}.$ It is readily seen that, on account of the last factor, we obtain $J_2=0$ for the second integral.

In order finally to calculate the third integral $J_3$ we shall lay off in the series of values of $\omega$ on both sides of ${\omega }_0$ an interval extending from ${\omega }_1$ ($<{\omega }_0$) to ${\omega }_2$ ($>{\omega }_0$) such that $\frac{{\omega }_0-{\omega }_1}{{\omega }_0}\text{and}\frac{{\omega }_2-{\omega }_0}{{\omega }_0}\text{are small,}\mathrm{\backslash Label}[eqn](243)\text{upshape (243)}$ and simultaneously $({\omega }_0-{\omega }_1)\tau \text{and}({\omega }_2-{\omega }_0)\tau \text{are large.}\mathrm{\backslash Label}[eqn](244)\text{upshape (244)}$ This can always be done, since ${\omega }_0\tau$ is large. If we now break up the integral $J_3$ into three parts, as follows: $J_3={\int }_0^{\infty }={\int }_0^{{\omega }_1}+{\int }_{{\omega }_1}^{{\omega }_2}+{\int }_{{\omega }_2}^{\infty },$ it is seen that in the first and third partial integral the expression ${\sin }^2{\displaystyle \frac{{\omega }_0-\omega }{2}}\tau$ may, because of the condition [eqn:(244)] (244), be replaced by its mean value $\frac{1}{2}$. Then the two partial integrals become: ${\int }_0^{{\omega }_1}\frac{{\omega }_{0}d\omega }{2({\omega }_0+\omega )({\omega }_0-\omega {)}^2}\text{and}{\int }_{{\omega }_2}^{\infty }\frac{{\omega }_{0}d\omega }{2({\omega }_0+\omega )({\omega }_0-\omega {)}^2}.\mathrm{\backslash Label}[eqn](245)\text{upshape (245)}$ These are certainly smaller than the integrals: ${\int }_0^{{\omega }_1}\frac{d\omega }{2({\omega }_0-\omega {)}^2}\text{and}{\int }_{{\omega }_2}^{\infty }\frac{d\omega }{2({\omega }_0-\omega {)}^2}$

which have the values $\frac{1}{2}\frac{{\omega }_1}{{\omega }_0({\omega }_0-{\omega }_1)}\text{and}\frac{1}{2({\omega }_2-{\omega }_0)}\mathrm{\backslash Label}[eqn](246)\text{upshape (246)}$ respectively. We must now consider the middle one of the three partial integrals: ${\int }_{{\omega }_1}^{{\omega }_2}d\omega \frac{{\omega }_0}{({\omega }_0+\omega )({\omega }_0-\omega {)}^2}·{\sin }^2\frac{{\omega }_0-\omega }{2}\tau .$ Because of condition [eqn:(243)] (243) we may write instead of this: ${\int }_{{\omega }_1}^{{\omega }_2}d\omega ·\frac{{\sin }^2{\displaystyle \frac{{\omega }_0-\omega }{2}}\tau }{2({\omega }_0-\omega {)}^2}$ and by introducing the variable of integration $x$, where $x=\frac{\omega -{\omega }_0}{2}\tau$ and taking account of condition [eqn:(244)] (244) for the limits of the integral, we get: $\frac{\tau }{4}{\int }_{-\infty }^{+\infty }\frac{{\sin }^{2}xdx}{x^2}=\frac{\tau }{4}\pi .$ This expression is of a higher order of magnitude than the expressions [eqn:(246)] (246) and hence of still higher order than the partial integrals [eqn:(245)] (245) and the integrals $J_1$ and $J_2$ given above. Thus for our calculation only those values of $\omega$ will contribute an appreciable part which lie in the interval between ${\omega }_1$ and ${\omega }_2$, and hence we may, because of [eqn:(243)] (243), replace the separate coefficients $A_n^2$ and $B_n^2$ in the expression for the total absorbed energy by their mean values $A_0^2$ and $B_0^2$ in the neighborhood of ${\omega }_0$ and thus, by taking account of [eqn:(242)] (242), we shall finally obtain for the total value of the energy absorbed by the oscillator in the time $\tau$: $\frac{1}{L}\frac{\tau }{8}(A_0^2+B_0^2)𝖳.\mathrm{\backslash Label}[eqn](247)\text{upshape (247)}$ If we now, as in [eqn:(158)] (158), define $𝖨$, the "intensity of the vibration

exciting the oscillator," by spectral resolution of the mean value of the square of the exciting field-strength ${⋿}_z$: $\overline{{⋿}_z^2\mathrm{\backslash Strut}}={\int }_0^{\infty }{𝖨}_{\nu }d\nu \mathrm{\backslash Label}[eqn](248)\text{upshape (248)}$ we obtain from [eqn:(235)] (235) and [eqn:(242)] (242): $\overline{{⋿}_z^2\mathrm{\backslash Strut}}=\frac{1}{2}{\sum }_1^{\infty }(A_n^2+B_n^2)=\frac{1}{2}{\int }_0^{\infty }(A_n^2+B_n^2)𝖳d\nu ,$ and by comparison with [eqn:(248)] (248): $𝖨=\frac{1}{2}(A_0^2+B_0^2)𝖳.$ Accordingly from [eqn:(247)] (247) the energy absorbed in the time $\tau$ becomes: $\frac{𝖨}{4L}\tau ,$

that is, in the time between two successive emissions, the energy $U$ of the oscillator increases uniformly with the time, according to the law $\frac{dU}{dt}=\frac{𝖨}{4L}=a.\mathrm{\backslash Label}[eqn](249)\text{upshape (249)}$ Hence the energy absorbed by all $N$ oscillators in the time $dt$ is: $\frac{N𝖨}{4L}dt=Nadt.\mathrm{\backslash Label}[eqn](250)\text{upshape (250)}$