I'm wrestling with this theorem in my 'stationary stochastic processes' course.It's about sampling of a continous process and a way of rewriting the covariance function of the sample(I'm not entirely sure so correct me if I'm wrong). Here they use $f_s$=1/d(which is the sampling frequency and $f_n$=1/(2d)(Nyquist frequency) and the proof goes:
"The spectrum of the sampled process {$Z_t$, t = $0, \pm d, \pm 2d$} can be concentrated in the interval ${-f_n}<f\le{f_n}$, where $f_n$=1/2d is the Nyquist frequency. The covariance function $r_Z(\tau)$ can be expressed as $r_Z(\tau)=\int_{{-f_n}+0}^{f_n} e^{i2{\pi}f\tau} R_Z(f)df$ for $\tau$ = 0,$\pm 1, \pm 2,..$ with a spectral density $R_z(f)$=$\sum\limits_{-\infty}^{\infty} R_Y(f+kf_s)$ for $-f_n<f\le f_n$.
$r_Z(\tau)$= $r_Y(nd)$= $\int_{-\infty}^{\infty}e^{i2{\pi}fnd}R_Y(f)df$= $\sum\limits_{-\infty}^{\infty}\int_{-(2k-1)f_n}^{(2k+1)f_n}e^{i2{\pi}fnd} R_Y(f)df$= $\\\sum\limits_{-\infty}^{\infty}\int_{-f_n}^{f_n}e^{i2{\pi}(f+kf_s)nd} R_Y(f+kf_s)df=\\$ $\int_{-f_n}^{f_n}\sum\limits_{-\infty}^{\infty}e^{i2{\pi}(f+kf_s)nd} R_Y(f+kf_s)df$
Since $e^{i2\pi(f+kf_s)nd}=e^{i2{\pi}fnd}$ this is equal to$\int_{-\infty}^{\infty}e^{i2{\pi}fnd}\{\sum_{k=-\infty}^{\infty}R_Y(f+kf_s)\}$ and hence $r_Z(\tau)$, with $\tau$=nd, can be expressed as a fourier transform of $R_Z(f)=\sum_{k=-\infty}^{\infty}R_Y(f+kf_s)$, with f in the interval (-$f_n,f_n]$."
I don't get what happends when they say that $\sum\limits_{-\infty}^{\infty}\int_{-(2k-1)f_n}^{(2k+1)f_n}e^{i2{\pi}fnd} R_Y(f)df=\sum\limits_{-\infty}^{\infty}\int_{-f_n}^{f_n}e^{i2{\pi}(f+kf_s)nd} R_Y(f+kf_s)df$ Could someone explain (and help me get some intuition on) what they're doing?
Thankfully,