relation between DFT to CTFT
Let's give it a shot. 

We know that during sampling, we obtained a discretized version, $x(nT_s)$ of a continuous time signal $x(t)$ sampled every $T_s$ seconds. I will denote their corresponding Fourier Transforms as $X_{DT}(e^{j\omega})$ and $X_{CT}(j\Omega)$, respectively, according to the majority of the bibliography. Indices denote discrete time (DT) and continuous-time (CT) Fourier Transforms. 

The Discrete time Fourier Transform (DTFT) of the sampled signal is given by $$X_{DT}(e^{j\omega}) = \frac{1}{T_s}\sum_{l=-\infty}^{+\infty}X_{CT}\Big(j\Big(\frac{\omega}{T_s} - l\frac{2\pi l}{T_s}\Big)\Big)$$ 

Assuming that there is no aliasing (like in your case), over one period this turns into $$X_{DT}(e^{j\omega}) = \frac{1}{T_s}X_{CT}\Big(j\frac{\omega}{T_s}\Big)$$

Due to the $2\pi$-periodic nature of the DTFT and the fact that we're dealing with real-valued signals, we can rewrite this as $$X_{DT}(e^{j\omega}) = \left\{\begin{array}{ll}
\frac{1}{T_s}X_{CT}\Big(j\frac{\omega}{T_s}\Big), &amp; 0 \leq \omega &lt; \pi \\
\frac{1}{T_s}X_{CT}\Big(j\frac{\omega-2\pi}{T_s}\Big), &amp; \pi \leq \omega &lt; 2\pi \\
\end{array}\right.$$

The Discrete Fourier Transform (DFT) is an $N-$point uniformly sampled version of the DTFT, so $$X_{DFT}[k] = X_{DT}(e^{j\omega})\Big|_{\omega = 2\pi k/N}$$ for $0 \leq k\leq N-1$.

Combining the last two equations we have $$X_{DFT}[k] = \left\{\begin{array}{ll}
\frac{1}{T_s}X_{CT}\Big(j\frac{2\pi k}{NT_s}\Big), &amp; 0 \leq k &lt; N/2 \\
\frac{1}{T_s}X_{CT}\Big(j\frac{2\pi (k-N)}{NT_s}\Big), &amp; N/2 \leq k \leq N-1 \\
\end{array}\right.$$

Except for the constant $N$ in your results - that I have failed to extract - I think what I wrote will help you.