91Ó°ÊÓ

The given equation is$f\left(x\right)=\{\begin{array}{l}\begin{array}{ll}\mathrm{x},& \left|\mathrm{x}\right|<1\end{array}\\ \begin{array}{ll}0,& \left|\mathrm{x}\right|>1\end{array}\end{array}$

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$g\left(\mathrm{Î\pm }\right)$.

$\begin{array}{c}g\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}}{∫}_{−1}^{1}x{e}^{−i\mathrm{Î\pm }x}dx\\ ={\frac{1}{2\mathrm{Ï€}}[−\frac{x}{i\mathrm{Î\pm }}{e}^{−i\mathrm{Î\pm }x}+\frac{1}{{\mathrm{Î\pm }}^2}{e}^{−i\mathrm{Î\pm }x}]|}_{−1}^1\\ =\frac{1}{2\mathrm{Ï€}}[−\frac{e^{−i\mathrm{Î\pm }}}{i\mathrm{Î\pm }}−\frac{e^{i\mathrm{Î\pm }}}{i\mathrm{Î\pm }}+\frac{1}{{\mathrm{Î\pm }}^2}(e^{−i\mathrm{Î\pm }}−e^{i\mathrm{Î\pm }})]\\ =\frac{1}{2\mathrm{Ï€}}\frac{−i2i\sin \mathrm{Î\pm }−2\mathrm{Î\pm }\cos \mathrm{Î\pm }}{{\mathrm{Î\pm }}^{2}i}\end{array}$

Further solving

_{$g\left(\mathrm{Î\pm }\right)=\frac{1}{\mathrm{Ï€}}\frac{\sin \mathrm{Î\pm }−\mathrm{Î\pm }\cos \mathrm{Î\pm }}{i{\mathrm{Î\pm }}^2}$}

Therefore, the exponential Fourier transform of the given equation is $f\left(x\right)=\frac{1}{\mathrm{Ï€}}{∫}_{−\mathrm{∞}}^{\mathrm{∞}}\frac{\sin \mathrm{Î\pm }−\mathrm{Î\pm }\cos \mathrm{Î\pm }}{i{\mathrm{Î\pm }}^2}{e}^{i\mathrm{Î\pm }x}d\mathrm{Î\pm }$.