Amplitude after Fourier transform
Relating the DFT to the FT (CTFT) is a big issue.  Let's start with the basic definitions without any domain specification using a $\frac{1}{N}$ normalization on the DFT.
$$ FT(x(t))(f) = \int x(t) e^{-i2\pi t f } dt  $$
$$ DFT(x[n])(k) = \frac{1}{N} \sum x[n] e^{-i 2\pi \frac{n}{N} k } $$
They are obviously similar.  Assuming they cover the same interval, we can map the domain variables to each other.
$$
\begin{aligned}
t &amp;= \frac{n}{N} = \frac{1}{N} \cdot n \\
f &amp;= k  \\
( \Delta t  &amp;=\frac{1}{N} ) \to dt
\end{aligned}
$$
[Edit: Does anybody want to argue the (1/N) doesn't belong in the DFT definition?]
The units of $f$ and $k$ are cycles per the interval.
Of course, we are assuming the sequence of values matches the function at every point.
$$ x[n] = x(t) $$
The FT is defined on a domian of $-\infty$ to $\infty$, and the the DFT is defined on a domain of $0$ to $N-1$.  The corresponding $t$ values are $0$ and $\frac{N-1}{N}$.  We can now include the domains in the definitions for a direct comparison on the same interval.
$$ FT_{DFT}(x(t))(f) = \int_{0}^{\frac{N-1}{N}} x(t) e^{-i2\pi t f } dt  $$
$$ DFT(x[n])(k) = \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-i 2\pi \frac{n}{N} k } $$
To make the $FT_{DFT}$ transform fit the $FT$ transform, we have to introduce a rectangular window function on the interval.
$$ w(t) =
\begin{cases}
= 0 &amp; t &lt; 0 \\
= 1 &amp; 0 \le t &lt; 1 \\
= 0 &amp; 1 \le t \\
\end{cases}
$$
$$ FT_{DFT}(x(t))(f) = \int_{-\infty}^{\infty} w(t) x(t) e^{-i2\pi t f } dt  $$
Please notice that there is no window function in the DFT! The window function is in the FT.
From these definitions, it is very clear that the FT is the limit of the DFT with a $\frac{1}{N}$ normalization factor as $N$ goes to infinity.
$$
\begin{aligned}
\lim_{N \to \infty} DFT(x[n]) &amp;= FT_{DFT}(x(t)) \\
\lim_{N \to \infty} \frac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-i 2\pi \frac{n}{N} k } &amp;= \int_{-\infty}^{\infty} w(t) x(t) e^{-i2\pi t f } dt \\
\end{aligned}
$$
This is the &quot;right&quot; way to view it.  An integral is the limit of a summation.
It is possible to to view the DFT as a sampled version of the FT by using a second window &quot;function&quot; (loosely speaking) of a train of Dirac deltas.  But this is not the definition of the DFT, requires advanced Real Analysis understanding, and you are merely getting back to the definition by working backwards.
Some EE professors neglect to mention this, some just don't know it.
So, just like you can approximate an intergral with a summation, you can approximate an FT on a finite interval with a $\frac{1}{N}$ normalized DFT.
To get intervals other than $[0,1)$ on $t$, call your new time $\tau$ and rescale it.
$$ t = \frac{\tau}{T} = \frac{n}{N} $$
Feel free to shift it too.
$$ t = \frac{\tau- \tau_0}{T} = \frac{n}{N} $$
This makes the sequence definition equivalent to
$$ x[n] = x_{\tau}(\tau) = x_{\tau}\left( \tau_0 + \frac{T}{N}n \right) $$
The DFT does not care what your sampling rate is. aka $\frac{N}{T}$ with these names nor where you place the frame, $\tau_0$.

Followup:
Questions from the OP's comment under the post question:
(All answers assume you are dealing with a pure real tone with a slighlty varying envelope.)
&quot;How do you get rid of the influence of the number of samples and the range over which y(x) is defined? &quot;
By framing your interval on a whole number of cycles.  This is the principal justification in my mind for a $\frac{1}{N}$ normalization factor of the DFT.  On a whole number of cycles (for a pure steady tone) a single conjugate pair of bins are non-zero.  The value in the bin is independent of N when normalized.  Thus the reading you get matches the continuous case exactly and is independent of the sample count.  Also, there is no need to calculate the whole DFT (take a FFT), a single bin calculation will do.  With a consistent frequency having a set of basis vectors is most efficient.  With a varying frequency, a Goertzel calculation yields the equivalent more efficiently.
&quot;How do you obtain the correct amplitude of the Fourier transform of the function y(x)? &quot;
With a single bin, the magnitude of the bin is half the amplitude of the tone.  A small DFT frame size of 2 to 3 cycles in length ensures that the reading is fairly accurate on that interval.  The reading will be that of the best fit sinusoid on that frame.
&quot;How do you properly take into account the number of samples and the range over which y(x) is defined? &quot;
Not sure what you mean by this.  The sample rate and signal frequency define how many samples per cycle you will have.  Decide the number of cycles you want, multiply that by the number of samples in a cycle and you get the sample count (N) for that frame.  If you are calculating a single bin, there is no special advantage to powers of two or any other factor.