Approximating Lorentzian Fourier transform with FFT
First of all, take a look at this very related question and its answer(s). Second, your analytic solution is wrong, so it's no surprise that you don't see the correct result.

The time domain function

$$x(t)=e^{j2\pi f_0t}e^{-Rt}\tag{1}$$

has the following Fourier transform:

$$X(f)=\frac{1}{R}\frac{1-j\frac{2\pi(f-f_0)}{R}}{1+\left(\frac{2\pi(f-f_0)}{R}\right)^2}\tag{2}$$

Also, scaling should be multiplicative and not additive (unless you're in the logarithmic domain), and you should also not only look at the real parts of complex-valued functions.

As explained in this answer, approximating the CTFT by the DFT usually results in two types of errors: the truncation error (due to truncation of the time domain function), and the aliasing error (due to sampling the time domain function). These errors can be made small by choosing a large sampling frequency and a large DFT length.

Below is an example code in Octave/Matlab showing how the CTFT of the given function can be approximated by the DFT:


F0 = 0.3;
R = 0.04;

Fs = 30;            % sampling frequency
Ts = 1/Fs;
Tmax = 100;         % length of time domain signal
N = round(Tmax/Ts); % DFT length

t = (0:N-1) * Tmax / (N-1);
N2 = round(N/2);
f = (0:N2-1)/N*Fs;

fun = exp(1i*2*pi*F0*t) .* exp(-R*t);

% analytic expression for CTFT
z = 2*pi*(f - F0) / R;
FTfun = (1 - 1i*z) ./ (1 + z.^2) / R;

% DFT approximation of CTFT
FTfun2 = Ts * fft(fun,N);
FTfun2 = FTfun2(1:N2);

plot(f,20*log10(abs(FTfun)),f,20*log10(abs(FTfun2)))
    title('magnitudes (dB)'), xlabel('f'), legend('CTFT','DFT'), grid on
    axis([0,Fs/2,-40,30])