I am using Digital Signal Processing Principles, Algorithms, and Applications 4th edition and by Proakis. Here is what I don't understand
say the signal $x_a(t)$ has its Fourier Transform $X_a(F)$ and digital signal $x(n)=x_a(nT)$ which is $x_a(t)$ after sampling with sampling frequency $F_s$ has its DTFT $X(f)$
since $$x_a(t)=\int^{\infty}_{-\infty}X_a(F)\,e^{j2\pi Ft}dF\tag{1}$$
because $x(n)=x_a(nT)$ and $nT=\dfrac{n}{F_s}$ $$x(n)=x_a(nT)=\int^{\infty}_{-\infty}X_a(F)e^{j2\pi nF/F_s}dF\tag{2}$$
Using inverse DTFT for $x(n)$ then
$$\int^{\frac12}_{-\frac12}X(f)e^{j2\pi nf}df=\int^{\infty}_{-\infty}X_a(F)e^{j2\pi nF/F_s}dF\tag{3}$$
for $f=\dfrac{F}{F_s}$ then
$$\int^{\frac12}_{-\frac12}X(f)e^{j2\pi nf}df=\frac{1}{F_s}\int^{\frac {F_s}{2}}_{-{\frac {F_s}{2}}}X(F)e^{j2\pi nF/F_s}dF\tag{4}$$
That is what the book says and it use that to explain the reconstruction process in theory. However isn't it will be like this according to substitution in integral
$$\int^{\frac12}_{-\frac12}X(f)e^{j2\pi nf}df = \frac{1}{F_s} \int^{\frac {F_s}{2}}_{-{\frac {F_s}{2}}}X\left(\frac{F}{F_s}\right)e^{j2\pi nF/F_s}dF\tag{5}$$
I think this should be in Math Section but since it is about Digital Signal so I post it here.