Why is $\sum_{m=1}^{n}\frac{(\nu(m)-\log\log n)^2}{n\log\log n}=\int_{-\infty}^{\infty}\omega^2\, \mathrm{d}\sigma_n(\omega)$?

Question

I was reading an article on Probabilistic Number Theory by M.Kac where I am not able to understand why a particular equation mentioned here in page $657$ equation $(7.7)$ is true?

I do understand that $\frac{(\nu(m)-\log\log n)^2}{\log\log n}<\omega^2$ and that it follows that $\frac{(\nu(m)-\log\log n)^2}{n\log\log n}<\frac{\omega^2}{n}$. But why after summing over $m$ we are getting a integral of $\omega^2$? Specifically, why did the inequality changed into an equality?

If somebody could explain this to me I would really appreciate it.

Answer 1

Denote $f_m=(\nu(m)-\log\log n)(\log\log n)^{-1/2}$ and define $\rho(\omega)=\sum_{m=1}^n\delta(\omega-f_m)$, with $\delta(x)$ the Dirac delta function. Because of the identity $\int g(\omega)\delta(\omega-\omega_0)\,d\omega=g(\omega_0)$, one has $$n^{-1}\sum_{m=1}^n f_m^2=n^{-1}\int \omega^2\rho(\omega)\,d\omega.$$

The function $\sigma(\omega)$ introduced by Kac is given by$^\ast$ $$\frac{d}{d\omega}\sigma(\omega)=n^{-1}\rho(\omega),$$ so one may equivalently write $$n^{-1}\sum_{m=1}^n f_m^2=\int \omega^2\,d\sigma(\omega),$$ which is the formula desired by the OP.

_{$^\ast$ The relation between $\sigma(\omega)$ and $\rho(\omega)$ follows from the identity $d\theta(\omega)/d\omega=\delta(\omega)$, with $\theta(\omega)$ the unit step function.}