How does the Fourier transform of the periodic extension relate to the Fourier series of ? (continuous signal)

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We know that in discrete signals, the relation between the Fourier transform of a signal and its Fourier series coefficient is as follows:

$$X(jkw_0) = 2\pi a_k$$

Or: $$X(jw) = 2\pi\sum_{k=0}^{k=N}{a_k\delta(w-kw_0)} $$

Which is provable very easily. But what is the relation between CT FS and CT FT? As we know if we have a $x_1(t)$ defined is:

$$x_1(t)=\left\{ \begin{array}{ll} f(t) & |t|< \frac{T_1}{2} \\ 0 & \mbox{oth.} \end{array} \right.$$

We can convolve it into a impulse train to create the periodic function $x_(t)$.

$$x(t) = x_1(t) * \sum_{k=-\infty}^{+\infty}\delta(t-kT)$$

And the fourier transform will be:

$$ X(jw) = X_1(jw) \times F(\sum_{k=-\infty}^{+\infty}\delta(t-kT)) $$

$$ X(jw) = X_1(jw) \times \sum_{k=-\infty}^{+\infty}e^{jwkT} $$

Which does not give any indication of any linear relation between CTFS and CTFT of a function.