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Given a square wave,
\[\begin{equation}f(t) = 1 \textrm{, for } 0 \leq t \leq t_0\end{equation}\]
The Fourier Transform is given by:
\[\begin{equation}F(\xi) = \int_{-\infty}^{\infty} f(t) e^{-i 2\pi \xi t} dt\end{equation}\]
Substituting the first equation to the second one yields:
\[\begin{equation}F(\xi) = \int_{0}^{t_0} e^{-i 2\pi \xi t} dt\end{equation}\]
\[\begin{equation}F(\xi) = -\frac{1}{i 2\pi \xi}\int_{0}^{t_0} e^{-i 2\pi \xi t} \cdot (-i 2\pi \xi) dt\end{equation}\]
\[\begin{equation}F(\xi) = -\frac{1}{i 2\pi \xi} \left[e^{-i 2\pi\xi t}\right]_0^{t_0}\end{equation}\]
\[\begin{equation}F(\xi) = -\frac{1}{i 2\pi \xi} \left[e^{-i 2\pi\xi t_0} - 1\right]\end{equation}\]
\[\begin{equation}F(\xi) = \frac{1}{i 2\pi \xi} \left[1 - e^{-i 2\pi\xi t_0} \right]\end{equation}\]
We can write this as:
\[\begin{equation}F(\xi) = \frac{1}{i 2\pi \xi}\frac{e^{i 2\pi\xi (t_0/2)}}{e^{i 2\pi\xi (t_0/2)}} \left[1 - e^{-i 2\pi\xi t_0} \right]\end{equation}\]
\[\begin{equation}F(\xi) = \frac{1}{\pi \xi}\frac{1}{e^{i 2\pi\xi (t_0/2)}} \left[\frac{e^{i 2\pi\xi (t_0/2)} - e^{-i 2\pi\xi (t_0/2)}}{2i} \right]\end{equation}\]
Thus,
\[\begin{equation}\boxed{F(\xi) = e^{-i\pi\xi t_0}\frac{\sin\left(\pi\xi t_0\right)}{\pi\xi}}\end{equation}\]
Substituting the first equation to the second one yields:
\[\begin{equation}F(\xi) = \int_{0}^{t_0} e^{-i 2\pi \xi t} dt\end{equation}\]
\[\begin{equation}F(\xi) = -\frac{1}{i 2\pi \xi}\int_{0}^{t_0} e^{-i 2\pi \xi t} \cdot (-i 2\pi \xi) dt\end{equation}\]
\[\begin{equation}F(\xi) = -\frac{1}{i 2\pi \xi} \left[e^{-i 2\pi\xi t}\right]_0^{t_0}\end{equation}\]
\[\begin{equation}F(\xi) = -\frac{1}{i 2\pi \xi} \left[e^{-i 2\pi\xi t_0} - 1\right]\end{equation}\]
\[\begin{equation}F(\xi) = \frac{1}{i 2\pi \xi} \left[1 - e^{-i 2\pi\xi t_0} \right]\end{equation}\]
We can write this as:
\[\begin{equation}F(\xi) = \frac{1}{i 2\pi \xi}\frac{e^{i 2\pi\xi (t_0/2)}}{e^{i 2\pi\xi (t_0/2)}} \left[1 - e^{-i 2\pi\xi t_0} \right]\end{equation}\]
\[\begin{equation}F(\xi) = \frac{1}{\pi \xi}\frac{1}{e^{i 2\pi\xi (t_0/2)}} \left[\frac{e^{i 2\pi\xi (t_0/2)} - e^{-i 2\pi\xi (t_0/2)}}{2i} \right]\end{equation}\]
Thus,
\[\begin{equation}\boxed{F(\xi) = e^{-i\pi\xi t_0}\frac{\sin\left(\pi\xi t_0\right)}{\pi\xi}}\end{equation}\]
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