The Fourier Transform(FT) is $X(\upsilon) = \int_{-\infty}^{\infty}x(t)e^{-2{\pi}i{\upsilon}t}dt$.
The impulse train is $\delta_1(x)=\sum\limits_{k=-\infty}^{\infty}\delta(x-k)$, and its FT is $\delta_1(\upsilon)$.
Sample $x(t)$ with $\delta_1(x)$, then the FT of $ x(t)\delta_1(x) $ is $\sum\limits_{k=-\infty}^{\infty}x(k)e^{-2\pi i\upsilon k}$.
On the other hand, the FT of $ x(t)\delta_1(x) $ is $X(\upsilon)\star\delta_1(\upsilon) = \sum\limits_{k=-\infty}^{\infty}X(\upsilon-k)$.
So $\sum\limits_{k=-\infty}^{\infty}x(k)e^{-2\pi i\upsilon k} = \sum\limits_{k=-\infty}^{\infty}X(\upsilon-k)$.
Now whether $x(t)$ is compactly supported on $[0, N]$ or $[0, N)$, $X(\upsilon)$ seems to be be the same, so does the right part $\sum\limits_{k=-\infty}^{\infty}X(\upsilon-k)$, but the left part $\sum\limits_{k=-\infty}^{\infty}x(k)e^{-2\pi i\upsilon k}$ differs. One is $\sum\limits_{k=0}^{N}x(k)e^{-2\pi i\upsilon k}$, and the other is $\sum\limits_{k=0}^{N-1}x(k)e^{-2\pi i\upsilon k}$.
So what is the condition for $\sum\limits_{k=-\infty}^{\infty}x(k)e^{-2\pi i\upsilon k} = \sum\limits_{k=-\infty}^{\infty}X(\upsilon-k)$ ?
I think it is right if $x(t)$ is continuous, but don't know why. And which book is recommended for this problem ?
This is known as the Poisson summation formula and related to Fourier series and the sampling theorem. It is sufficient if both the function and its Fourier transform fall faster than quadratically. If I remember correctly (yes, see wikipedia), the boundedness of
$$(1+|t|)^{1+\varepsilon}x(t)\text{ and }(1+|v|)^{1+\varepsilon}\,|X(v)|$$
for some $\varepsilon>0$ is a rather minimal sufficient condition.
All this is mathematically expanded on nicely in Benedetto/Zimmermann 1997, also to recommend as nice reading is Higgins: Five stories... 1985.