Theorem:
$$
X(\omega)=\int_{t=-\infty}^{\infty}x(t) e^{-i \omega t} dt $$
$$
x(t)=\frac{1}{2\pi}\int_{\omega=-\infty}^{\infty}X(\omega) e^{i \omega t} d\omega $$
EXAMPLES
Example 1: Find the FT of $x(t)=e^{-\alpha t}U(t)$ for $\alpha>0$.
Answer 1:
$$\begin{eqnarray}
X(\omega)&=&\int_{t=-\infty}^{\infty}e^{-\alpha t}U(t) e^{-i \omega t} dt\\
&=&\int_{t=0}^{\infty}e^{-(\alpha t+i \omega) t} dt\\
&=&\left. \frac{e^{-(\alpha t+i \omega) t}}{-(\alpha +i \omega)}\right\vert_{t=0}^{\infty}\\
&=&\frac{1}{\alpha +i \omega}
\end{eqnarray} $$
Example 2: Find the FT of $x(t)=U(t)$.
Answer 2:
$$\begin{eqnarray}
X(\omega)&=&\int_{t=-\infty}^{\infty}U(t) e^{-i \omega t} dt\\
&=&\int_{t=0}^{\infty}e^{-i \omega t} dt\\
&=&\left. \frac{e^{-i \omega t}}{-i \omega}\right\vert_{t=0}^{\infty}\\
&=&\frac{1}{i \omega}-\lim_{t \rightarrow \infty }\frac{e^{-i \omega t}}{i \omega}\\
&=&\frac{1}{i \omega}+\pi\delta(\omega)
\end{eqnarray} $$
Example 3: Find the FT of $x(t)=1$.
Answer 3:
$$\begin{eqnarray}
X(\omega)&=&\int_{t=-\infty}^{\infty}1. e^{-i \omega t} dt\\
&=&\left. \frac{e^{-i \omega t}}{-i \omega}\right\vert_{t=-\infty}^{\infty}\\
&=&\lim_{t \rightarrow \infty }\frac{e^{i \omega t}-e^{-i \omega t}}{i \omega}\\
&=&\lim_{t \rightarrow \infty }2\frac{\sin(\omega t)}{\omega}\\
&=& 2 \pi \delta(\omega)
\end{eqnarray} $$
Example 4: Find the FT of $\chi_{[-T,T]}(t)$.
Answer 4:
$$\begin{eqnarray}
X(\omega)&=&\int_{t=-\infty}^{\infty}\chi_{[-T,T]}(t) e^{-i \omega t} dt\\
&=&\int_{t=-T}^{T} e^{-i \omega t} dt\\
&=&\left. \frac{e^{-i \omega t}}{-i \omega}\right\vert_{t=-T}^{T}\\
&=&\frac{e^{i \omega T}-e^{-i \omega T}}{i \omega}\\
&=&2\frac{\sin(\omega T)}{\omega}
\end{eqnarray} $$
Note that Example 3 is a special case of this example, as $T \rightarrow \infty$:
$$\begin{eqnarray}
\lim_{T \rightarrow \infty }\chi_{[-T,T]}(t) &\rightarrow& 1\\
\lim_{T \rightarrow \infty } 2\frac{\sin(\omega T)}{\omega} &\rightarrow& 2\pi \delta(\omega)
\end{eqnarray} $$
Example 5: Find the FT of $x(t)=e^{iat}$ .
Answer 5:
$$\begin{eqnarray}
X(\omega)&=&\int_{t=-\infty}^{\infty}e^{iat}. e^{-i \omega t} dt\\
&=&\left. \frac{e^{i (a-\omega) t}}{i (a-\omega)}\right\vert_{t=-\infty}^{\infty}\\
&=&\lim_{t \rightarrow \infty }\frac{e^{i (a-\omega) t}-e^{-i (a-\omega) t}}{i (a-\omega)}\\
&=&\lim_{t \rightarrow \infty }2\frac{\sin((a-\omega) t)}{a-\omega}\\
&=& 2 \pi \delta(a-\omega)
\end{eqnarray} $$
Note that example 3 is a special case of this example for $a=0$. (also remember that $\delta(t)=\delta(-t))$
Example 6: Find the FT of $x(t)=\delta(t-a)$.
Answer 6:
$$\begin{eqnarray}
X(\omega)&=&\int_{t=-\infty}^{\infty}\delta(t-a) e^{-i \omega t} dt\\\\
&=& e^{-i a t}
\end{eqnarray} $$
Example 7: Find the FT of $x(t)=e^{-a|t|}$ for $a>0$.
Answer 7:
$$\begin{eqnarray}
X(\omega)&=&\int_{t=-\infty}^{\infty}e^{-a|t|} e^{-i \omega t} dt\\\\
&=&\int_{t=-\infty}^{0}e^{at} e^{-i \omega t} dt+\int_{t=0}^{\infty}e^{-at} e^{-i \omega t} dt\\\\
&=&\left. \frac{e^{(\alpha -i \omega) t}}{\alpha -i \omega}\right\vert_{t=-\infty}^{0} + \left. \frac{e^{(-\alpha -i \omega) t}}{-\alpha -i \omega}\right\vert_{t=0}^{\infty}\\\\
&=&\frac{1}{\alpha -i \omega}+\frac{1}{\alpha +i \omega}\\\\
&=&\frac{2a}{\alpha^2 + \omega^2}
\end{eqnarray} $$
SUM OF COMPLEX EXPONENTIALS
Consider the geometric series
$$\begin{eqnarray}
\sum_{n=-k}^{k} e^{-i A n} &=& e^{-i A k}+e^{-i A (k-1)}+\ldots+e^{-i A (k-1)}+ e^{-i A k}\\
&=& e^{-i A k} \left(1+e^{i A }+e^{i 2A}+\ldots+e^{i (2k-1)A}+ e^{i 2kA}\right)\\
&=&e^{-i kA} \frac{1-e^{(2k+1)Ai }}{1-e^{i A}}
\end{eqnarray} $$
To ease the notation, let us define $r=e^{i A}$. Then, the above expression can be simplified as
$$\begin{eqnarray}
r^{-k}\frac{1-r^{2k+1}}{1-r}
&=& r^{-k} \frac{r^\frac{{2k+1}}{2}\left[r^{-\frac{{2k+1}}{2}}-r^\frac{{2k+1}}{2}\right]}{r^{\frac{1}{2}}\left[ r^{-\frac{1}{2}}-r^{\frac{1}{2}}\right]}\\
&=&\frac{r^{-\frac{{2k+1}}{2}}-r^\frac{{2k+1}}{2}}{ r^{-\frac{1}{2}}-r^{\frac{1}{2}}}
\end{eqnarray} $$
Substituting back $r=e^{-i \omega T}$,
$$\begin{eqnarray}
&=&\frac{e^{\frac{2k+1}{2}i \omega T}-e^{\frac{-{2k+1}}{2}i \omega T}}{ e^{\frac{i \omega T}{2}}-e^{-\frac{i \omega T}{2}}}\\\\
&=&\frac{\frac{e^{\frac{2k+1}{2}i \omega T}-e^{\frac{-{2k+1}}{2}i \omega T}}{2i}}{ \frac{e^{\frac{i \omega T}{2}}-e^{-\frac{i \omega T}{2}}}{2i}}\\\\
&=& \frac{\sin\left(\frac{2k+1}{2} \omega T\right)}{\sin\left(\frac{ \omega T}{2}\right)}
\end{eqnarray} $$
Or, in summary, we have proved that
$$\begin{eqnarray}
\sum_{n=-k}^{k} e^{-i \omega nT} &=& \frac{\sin\left(\frac{2k+1}{2} \omega T\right)}{\sin\left(\frac{ \omega T}{2}\right)}
\end{eqnarray} $$
CFT OF IMPULSE TRAIN
Impulse train is a periodic signal with period $T$:
$$\begin{eqnarray}
x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)
\end{eqnarray} $$
Its CFT is
$$\begin{eqnarray}
X(\omega)&=&\int_{t=-\infty}^{\infty}\left( \sum_{k=-\infty}^{\infty}\delta(t-kT) \right)e^{-i \omega t} dt\\\\
&=& \sum_{k=-\infty}^{\infty}\int_{t=-\infty}^{\infty}\delta(t-kT) e^{-i \omega t} dt\\\\
&=& \sum_{k=-\infty}^{\infty}e^{-i \omega kT}
\end{eqnarray} $$
We can express this result as
$$\begin{eqnarray}
X(\omega)&=&\lim_{N \rightarrow \infty}\sum_{k=-N}^{N}e^{-i \omega kT}
\end{eqnarray} $$
Then, we can utilize the summation of complex exponentials formula given above, with the substitution $\omega T \Rightarrow A$.
$$\begin{eqnarray}
X(\omega)&=& \frac{\sin\left(\frac{2N+1}{2} \omega T\right)}{\sin\left(\frac{ \omega T}{2}\right)}
\end{eqnarray} $$
CFT OF PERIODIC FUNCTIONS
Let $x(t)=x(t+T)$. Then,
$$\begin{eqnarray}
X(\omega)&=&\int_{t=-\infty}^{\infty}x(t) e^{-i \omega t} dt \\
&=& \sum_{n=-\infty}^{\infty} \int_{t=nT}^{(n+1)T} x(t) e^{-i \omega t} dt\\
&=& \sum_{n=-\infty}^{\infty} \int_{t=0}^{T} x(t+nT) e^{-i \omega (t+nT)} dt\\
&=& \sum_{n=-\infty}^{\infty} \int_{t=0}^{T} x(t) e^{-i \omega (t+nT)} dt\\
&=& \sum_{n=-\infty}^{\infty} \int_{t=0}^{T} x(t) e^{-i \omega t} e^{-i \omega nT}dt\\
&=&\left( \int_{t=0}^{T} x(t) e^{-i \omega t}dt\right) \left( \sum_{n=-\infty}^{\infty} e^{-i \omega nT} \right)
\end{eqnarray} $$
We have to evaluate the left-hand paranthesis. We write it as
$$
\sum_{n=-\infty}^{\infty} e^{-i \omega nT} = \lim_{k \rightarrow \infty }\sum_{n=-k}^{k} e^{-i \omega nT}
$$
This is a geometric series with the sum
$$\begin{eqnarray}
\sum_{n=-k}^{k} e^{-i \omega nT} &=& e^{-i \omega kT}+e^{-i \omega (k-1)T}+\ldots+e^{-i \omega (k-1)T}+ e^{-i \omega kT}\\
&=& e^{-i \omega kT} \left(1+e^{i \omega T}+e^{i \omega 2T}+\ldots+e^{i \omega (2k-1)T}+ e^{i \omega 2kT}\right)\\
&=&e^{-i \omega kT} \frac{1-e^{(2k+1)i \omega T}}{1-e^{i \omega T}}
\end{eqnarray} $$
To ease the notation, let us define $r=e^{-i \omega T}$. Then, the above expression can be simplified as
$$\begin{eqnarray}
r^{-k}\frac{1-r^{2k+1}}{1-r}
&=& r^{-k} \frac{r^\frac{{2k+1}}{2}\left[r^{-\frac{{2k+1}}{2}}-r^\frac{{2k+1}}{2}\right]}{r^{\frac{1}{2}}\left[ r^{-\frac{1}{2}}-r^{\frac{1}{2}}\right]}\\
&=&\frac{r^{-\frac{{2k+1}}{2}}-r^\frac{{2k+1}}{2}}{ r^{-\frac{1}{2}}-r^{\frac{1}{2}}}
\end{eqnarray} $$
Substituting back $r=e^{-i \omega T}$,
$$\begin{eqnarray}
&=&\frac{e^{\frac{2k+1}{2}i \omega T}-e^{\frac{-{2k+1}}{2}i \omega T}}{ e^{\frac{i \omega T}{2}}-e^{-\frac{i \omega T}{2}}}\\\\
&=&\frac{\frac{e^{\frac{2k+1}{2}i \omega T}-e^{\frac{-{2k+1}}{2}i \omega T}}{2i}}{ \frac{e^{\frac{i \omega T}{2}}-e^{-\frac{i \omega T}{2}}}{2i}}\\\\
&=& \frac{\sin\left(\frac{2k+1}{2} \omega T\right)}{\sin\left(\frac{ \omega T}{2}\right)}
\end{eqnarray} $$
Or, in summary, we have proved that
$$\begin{eqnarray}
\sum_{n=-k}^{k} e^{-i \omega nT} &=& \frac{\sin\left(\frac{2k+1}{2} \omega T\right)}{\sin\left(\frac{ \omega T}{2}\right)}
\end{eqnarray} $$