91Ó°ÊÓ

The given function is $f\left(x\right)=\left\{\begin{array}{l}\sin x,\left|x\right|<\mathrm{Ï€}/2\\ 0,\left|x\right|>\mathrm{Ï€}/2\end{array}\right.$. The exponential Fourier transform of the function is to be found and the function is to be written as a Fourier integral.

The following are the formulas for the Fourier series transforms,

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

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

Use the formula$g\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}}\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}f\left(x\right)e^{−i\mathrm{Î\pm }x}dx$to find the Fourier transform.

$\begin{array}{c}g\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}}\underset{−\mathrm{Ï€}/2}{\overset{\mathrm{Ï€}/2}{∫}}\sin x{e}^{−i\mathrm{Î\pm }x}dx\\ =\frac{1}{4\mathrm{Ï€}i}\underset{−\mathrm{Ï€}/2}{\overset{\mathrm{Ï€}/2}{∫}}(e^{ix}−e^{−ix})e^{−i\mathrm{Î\pm }x}dx\\ =−\frac{i}{4\mathrm{Ï€}}\underset{−\mathrm{Ï€}/2}{\overset{\mathrm{Ï€}/2}{∫}}(e^{ix(1−\mathrm{Î\pm })}−e^{−ix(1+\mathrm{Î\pm })})dx\\ =−{\frac{i}{4\mathrm{Ï€}}[\frac{e^{ix(1−\mathrm{Î\pm })}}{i(1−\mathrm{Î\pm })}+\frac{e^{−ix(1+\mathrm{Î\pm })}}{i(1+\mathrm{Î\pm })}]|}_{−\mathrm{Ï€}/2}^{\mathrm{Ï€}/2}\end{array}$

Solve further to obtain$g\left(\mathrm{Î\pm }\right)$.

$\begin{array}{c}g\left(\mathrm{Î\pm }\right)=−\frac{i}{2\mathrm{Ï€}}[\frac{\sin (\mathrm{Ï€}(1−\mathrm{Î\pm })/2)}{1−\mathrm{Î\pm }}−\frac{\sin (\mathrm{Ï€}(1+\mathrm{Î\pm })/2)}{1+\mathrm{Î\pm }}]\\ =−\frac{i}{2\mathrm{Ï€}}[\frac{\cos (\mathrm{Ï€}\mathrm{Î\pm }/2)}{1−\mathrm{Î\pm }}−\frac{\cos (\mathrm{Ï€}\mathrm{Î\pm }/2)}{1+\mathrm{Î\pm }}]\\ =\frac{−i}{\mathrm{Ï€}}\frac{\mathrm{Î\pm }}{1−{\mathrm{Î\pm }}^2}\cos \left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)\end{array}$

Now, write f(x) as a Fourier integral using the formula $f\left(x\right)=\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}g\left(\mathrm{Î\pm }\right)e^{i\mathrm{Î\pm }x}d\mathrm{Î\pm }$

$f\left(x\right)=\frac{−i}{\mathrm{Ï€}}\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\frac{\mathrm{Î\pm }}{1−{\mathrm{Î\pm }}^2}\text{cos}\left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)e^{i\mathrm{Î\pm }x}d\mathrm{Î\pm }$.