91Ó°ÊÓ

The given function is $f\left(x\right)=\left\{\begin{array}{l}\text{cos}x,−\raisebox{1ex}{${\displaystyle \mathrm{π}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 2}$}\right.<x<\raisebox{1ex}{${\displaystyle \mathrm{π}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 2}$}\right.\\ 0,\left|x\right|>\raisebox{1ex}{${\displaystyle \mathrm{π}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 2}$}\right.\end{array}\right.$. 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(x).

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.

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

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

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

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

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