DFT of Transient vs. Analytical Solution
That's the case. Assuming you have a band-limited continuous-time signal $x_c(t)$, then the relation between its (analytically derived) CTFT, $X_c(j\Omega)$, and the (numerically computed) DTFT, $X_d(e^{j\omega})$, (or actually its DFT $X_d[k]$ ) of the samples, $x_d[n] = x_c(n T_s)$, of it taken at a period $T_s$ will be :

$$ X_d(e^{j\omega}) = \frac{1}{T_s} X_c( \frac{\omega}{T_s}) ~~~, |\omega| &lt; \pi $$ 

In practice you are computing the N-point DFT $X[k]$ of $x_d[n]$ which is the uniform samples of $X_d(e^{j \omega})|_{\omega = \frac{2\pi}{N} k}$, for $k=0,1,...,N-1$. And they have the same magnitude relation too.

$$ X[k] =X_d(e^{j\frac{2\pi}{N} k}) = \frac{1}{T_s} X_c( \frac{\omega}{T_s}) ~~~, |\omega| &lt; \pi $$ 

In other words, you will have a weight of 
$$ K = \frac{1}{T_s}$$

between the computed DFT / DTFT and the theoretical CTFT of the signal.