Reconstruction of Contiuous - Time Signals In terms of analog signals, we can represent digital signal as :

$$
x[n] \triangleq x_{a}(nT) = \int_{-\infty}^{\infty}X_{a}(f) \, e^{j2\pi f nT} \ \mathrm{d}f
$$

While if we focused on the integral on the right side and according to the Digital Signal Processing by John Proakis, chapter 6.1, we can rewrite it into :

$$
\int_{-\infty}^{\infty}X_{a}(f) \, e^{j2\pi f nT} \ \mathrm{d}f = \sum_{k=-\infty}^{\infty} \int_{(k-1/2)F_{s}}^{(k+1/2)F_{s}}X_{a}(f) \, e^{j2\pi nf/F_{s}}  \ \mathrm{d}f 
$$

where $ F_{s} \triangleq \frac1T $.
My question is how the second equation comes up ? what does the interval of $(k-1/2)F_{s}$ to $(k+1/2)F_{s}$ means ?

Furthermore, it is stated in the book that "observing the  $X_{a}(f)$ in the interval of  $(k-1/2)F_{s}$ to $(k+1/2)F_{s}$ is identical to $X_{a}(f-kF_{s})$ in the interval of $-F_{s}/2$ to $F_{s}/2$". Is there any explanation or derivation how both things are identical ?

Thank you so much, hope that my question is clear enough
Because sampled signals Spectrum will have copies of original spectrum at multiples of $f=F_s$. Even if aliasing happened due to choice of $F_s$, the spectrum of sampled signal $x(n.T_s)$ will be periodic in frequency always with period $F_s$. And you can take any one such $k^{th}$ copy or period and integrate from $\frac{k-1}{2}F_s$ to $\frac{k+1}{2}F_s$. Now if you vary $k$ from $-\infty$ to $\infty$, then it is equivalent to integration on LHS. 

Key: Sampling at $F_s$ will create a spectrum which is periodic in frequency with period $F_s$.