91Ó°ÊÓ

The given equation is$f\left(x\right)=\{\begin{array}{l}\begin{array}{ll}1,& \mathrm{Ï€}/2<\left|\mathrm{x}\right|<\mathrm{Ï€}\end{array}\\ \begin{array}{ll}0,& \text{otherwise}\end{array}\end{array}$

A Fourier series is an infinite sum of sines and cosines expansion of a periodic function. The orthogonality relationships of the sine and cosine functions are used in the Fourier Series.

The following are the formulas for the Fourier series transforms,

$\begin{array}{c}f\left(x\right)=\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}g\left(\mathrm{Î\pm }\right)e^{i\mathrm{Î\pm }x}d\mathrm{Î\pm }\\ g\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}}\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}f\left(x\right)e^{−i\mathrm{Î\pm }x}dx\end{array}$

Here $g\left(\mathrm{Î\pm }\right)$is called the Fourier transform of f(x).

Find the value of$g\left(\mathrm{Î\pm }\right)$.

$\begin{array}{c}g\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}}{∫}_{−\mathrm{Ï€}}^{−\mathrm{Ï€}/2}{e}^{−i\mathrm{Î\pm }x}dx+\frac{1}{2\mathrm{Ï€}}{∫}_{\mathrm{Ï€}/2}^{\mathrm{Ï€}}{e}^{−i\mathrm{Î\pm }x}dx\\ ={\frac{1}{2\mathrm{Ï€}}\frac{−1}{i\mathrm{Î\pm }}{e}^{−i\mathrm{Î\pm }x}|}_{−\mathrm{Ï€}}^{−\mathrm{Ï€}/2}−{\frac{1}{2\mathrm{Ï€}}\frac{1}{i\mathrm{Î\pm }}{e}^{−i\mathrm{Î\pm }x}|}_{\mathrm{Ï€}/2}^{\mathrm{Ï€}}\\ =\frac{i}{2\mathrm{Ï€}\mathrm{Î\pm }}(e^{i\mathrm{Î\pm }\mathrm{Ï€}/2}−e^{i\mathrm{Î\pm }\mathrm{Ï€}})+\frac{i}{2\mathrm{Ï€}\mathrm{Î\pm }}(e^{−i\mathrm{Î\pm }\mathrm{Ï€}}−e^{−i\mathrm{Î\pm }\mathrm{Ï€}/2})\\ =\frac{i}{\mathrm{Ï€}\mathrm{Î\pm }}\left(i\sin \frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}−i\sin \left(\mathrm{Î\pm }\mathrm{Ï€}\right)\right)\end{array}$

Further solving

_{$g\left(\mathrm{Î\pm }\right)=\frac{\sin \left(\mathrm{Î\pm }\mathrm{Ï€}\right)−\sin (\mathrm{Î\pm }\mathrm{Ï€}/2)}{\mathrm{Ï€}\mathrm{Î\pm }}$}

Therefore, the exponential Fourier transform of the given equation is $f\left(x\right)={∫}_{−\mathrm{∞}}^{\mathrm{∞}}\frac{\sin \left(\mathrm{Î\pm }\mathrm{Ï€}\right)−\sin (\mathrm{Î\pm }\mathrm{Ï€}/2)}{\mathrm{Î\pm }\mathrm{Ï€}}{e}^{i\mathrm{Î\pm }x}d\mathrm{Î\pm }$.