I'll briefly explain this problem I faced.
Let's take this simple signal: $$x(n)=\cos(\pi n)$$ The signal is not absolutely summable, however we can define its DTFT in terms of distributions. That is: $$X(e^{j\omega})=2\pi \delta(\omega+\pi+2\pi k)$$ Note that the impulse has amplitude equals $2\pi$.
Now, let's take this signal: \begin{equation} y(n)= \begin{cases}1 &n=0\\-1 &n=1\\0 & \text{elsewhere} \end{cases} \end{equation} and compute the DTFT of this signal according to this definition: $$X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x(n)e^{-j\omega n}$$ We obtain: $$Y(e^{j\omega})=1-e^{-j\omega}$$ Computation of the DFT of the signal returns: $$Y[0]=0\qquad Y[1]=2$$ so, ok, DFT is the sampled version of the DTFT because $Y(e^{j\omega})$ evaluated in $\omega=0$ and $\omega=\pi$ turns out to be exactly the values calculated from the DFT.
Anyway, how can I explain the amplification by $2\pi$ in $X(e^{j\omega})$? Note that using the following definition of DTFT the problem doesn't appear: $$X(e^{j2\pi f})=\sum_{n=-\infty}^{\infty}x(n)e^{-j2\pi f}$$ because the transform of $x(n)$ in terms of distributions is: $$ X(e^{j2\pi f})=\delta(f+0.5+k) $$ which has amplitude 1. Moreover: $Y(e^{j2\pi f})=1-e^{-j2\pi f}$.
So, is the definition of DTFT using $\omega$ wrong?
