Remark:by $\eqref{Fourier6}$
\[X^{*}(\omega)=(\int_{-\infty}^{\infty}x(t) e^{-jk\omega_0 t}dt)^*=\int_{-\infty}^{\infty}x(t) e^{jk\omega_0 t}dt=X(-\omega)\]If $x(t)$ is $T_0$-periodic, then its Fourier transform is
\[X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t} dt\]which is not exist in the general cases. However, if we consider Fourier series expression $x(t)$, which is
\[x(t)=\sum_{k=-\infty}^{\infty}X_k e^{jk\omega_0 t}\]and look at the fact that when $x(t)=e^{j\omega_0 t}$, its Fourier transformation $X(\omega)$
\[X(\omega)=\int_{-\infty}^{\infty}e^{j\omega_0 t}e^{-j\omega t} dt =2\pi\delta(\omega_0-\omega)=2\pi\delta(\omega-\omega_0)\]Therefore, it is reasonable to consider
\[\begin{align} X(\omega) &=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt \\ &=\int_{-\infty}^{\infty}\sum_{k=-\infty}^{\infty}X_k e^{jk\omega_0 t} e^{-j\omega t} dt \\ &=\sum_{k=-\infty}^{\infty}X_k \int_{-\infty}^{\infty}e^{jk\omega_0 t}e^{-j\omega t} dt \\ &=\sum_{k=-\infty}^{\infty}2\pi X_k\delta(\omega-k\omega_0) \label{FourierSeries_X} \end{align}\]Example:
\[x(t)=\sum_{k=-\infty}^{\infty}\delta(t-k)\]We directly compute $X(\omega)$ as following:
\[\begin{aligned} X(\omega)&=\int_{-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\delta(t-k)e^{-j\omega t}dt \\ &=\sum_{k=-\infty}^{\infty}\int_{-\infty}^{\infty}\delta(t-k)e^{-j\omega t}dt \\ &=\sum_{k=-\infty}^{\infty}e^{-j\omega k} \end{aligned}\]Other approaches is to get the Fourier Series Data of $x(t)$. First, observing that $x(t)$ has period $1$, which implies $\omega_0=2\pi$.
According $\eqref{Fourier2}$,
\[\begin{aligned} X_n&=\int_{-1/2}^{1/2} \sum_{k=-\infty}^{\infty} \delta(t-k) e^{-j n \omega t} dt\\ &=\int_{-1/2}^{1/2} \delta(t) e^{-j n \omega t} dt \quad \text{ (since only }\delta(t-0)\text{ has nonzero value in [-1/2,1/2] )} \\ &=\int_{-\infty}^{\infty} \delta(t) e^{-j n \omega t} dt \\ &=e^{-j n \omega 0}=1 \end{aligned}\]Finally, the Fourier seriers data of $x(t)$ is
\[x(t)=\sum_{k=-\infty}^{\infty}e^{-jk2\pi t}\]We have compute the Fourier transform of Fourier series in general, so by $\eqref{FourierSeries_X}$,
\[X(\omega)=\sum_{k=-\infty}^{\infty}2\pi\delta(\omega-k2\pi)\]We prefer the last method one.
Example: by integration property,
\[F(u(t))=\frac{1}{jw}+\pi\delta(\omega)\]Example:
\[x(t)=[1+km(t)]cos(\omega_c t)\]where $cos(\omega_c t)$ is called carrier signal, $m(t)$ is called message signal and $k$ is called the modulation index. Assume message singal is banedlimiited by $\omega_m \ll \omega_c$
We conclude that
\[X(\omega)=\pi(\delta(\omega-\omega_c)+\delta(\omega+\omega_c))+\frac{k}{2} (M(\omega-\omega_c)+M(\omega+\omega_c))\]If $x(t)$ has Fourier Transform $X(\omega)$
\[F(X(t))=2\pi x(-\omega)\]