Relation between the DTFT and CTFT in sampling- sample period isn't as the impulse train period
We are given two signals:

$$x_c(t)=\sum_{n=-\infty}^{\infty}x(nT_1)\delta(t-nT_2)\tag{1}$$

and

$$x_d[n]=x(nT_1)\tag{2}$$

where $x(t)$ is some known signal whose Fourier transform $X(j\Omega)$ exists, and $T_1&gt;0$ and $T_2&gt;0$, but generally $T_1\neq T_2$.

We want to express the DTFT $X_d(e^{j\omega})$ of $x_d[n]$ in terms of the CTFT $X_c(j\Omega)$ of $x_c(t)$. The DTFT $X_d(e^{j\omega})$ is an aliased version of $X(j\Omega)$:

$$X_d(e^{j\omega})=\frac{1}{T_1}\sum_{k=-\infty}^{\infty}X\left[j\left(\frac{\omega-2\pi k}{T_1}\right)\right]\tag{3}$$

In order to compute the CTFT $X_c(j\Omega)$ of $x_c(t)$ we define a compressed ($T_1&gt;T_2$) or expanded ($T_1&lt;T_2$) version of $x(t)$:

$$\tilde{x}(t)=x\left(\frac{T_1}{T_2}t\right)\tag{4}$$

The CTFT of $\tilde{x}(t)$ is

$$\tilde{X}(j\Omega)=\frac{T_2}{T_1}X\left(j\frac{T_2}{T_1}\Omega\right)\tag{5}$$

With $(4)$ we have $x(nT_1)=\tilde{x}(nT_2)$, and $x_c(t)$ can be written as

$$x_c(t)=\sum_{n=-\infty}^{\infty}\tilde{x}(nT_2)\delta(t-nT_2)\tag{6}$$

The CTFT of $(6)$ is

$$X_c(j\Omega)=\frac{1}{T_2}\sum_{k=-\infty}^{\infty}\tilde{X}\left[j\left(\Omega-\frac{2\pi k}{T_2}\right)\right]\tag{7}$$

and with $(5)$ this becomes

$$X_c(j\Omega)=\frac{1}{T_1}\sum_{k=-\infty}^{\infty}X\left[j\left(\frac{T_2}{T_1}\Omega-\frac{2\pi k}{T_1}\right)\right]\tag{8}$$

Comparing $(3)$ and $(8)$ we finally obtain

$$X_d(e^{j\omega})=X_c\left(j\frac{\omega}{T_2}\right)\tag{9}$$