We know that sampling a continuous signal and converting it into discrete signal can be represented by timing an impulse train on the original signal and then scaling at time axis. The procedure described above in frequency domain is like the figure below, where the $T$ is the sampling period, $\omega_{M}$ is the maximum frequency component in original signal.
sampling procedure in frequency domain
We can see that the spectrum has a scaling factor $\frac{1}{T}$ here.
Now if the continuous signal is $cos(20t)$ and the sampling period $T$ is $\frac{\pi}{50}$, the sampled discrete signal $x_s[n]$ can be expressed as
$$ x(t) = cos(20t) $$
$$ x_s[n] = x(nT) = cos(20 \cdot n\frac{\pi}{50}) = cos(\frac{2\pi}{5}n) $$
Let the Fourier Transform of original signal is denoted by $X(j\omega)$ and the Fourier Transform of sampled discrete signal is denoted by $X_s(e^{j\omega})$. By looking at the FT table, their expression are $$ X(j\omega) = \pi\{\delta(\omega - 20) + \delta(\omega + 20)\}$$ and $$ X_{s}(e^{j\omega}) = \pi\sum_{l=-\infty}^{\infty}\{\delta(\omega-\frac{2\pi}{5} - 2\pi l) + \delta(\omega+\frac{2\pi}{5} - 2\pi l)\}$$
My problem is where does the scaling factor $\frac{1}{T}$ in $X_{s}(e^{j\omega})$ go?