Find the exponential Fourier transform of the given f(x)and write f(x)as a Fourier integral.

$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}\mathrm{cosx},-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.<x<\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 0,\left|x\right|>\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\end{array}$

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

$\begin{array}{c}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\underset{-\infty }{\overset{\infty }{\int }}g\left(\alpha \right)e^{i\alpha x}d\alpha \\ \mathbf{g}\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)e^{-i\alpha x}dx\end{array}$

Here $g\left(\alpha \right)$is called the Fourier transform of f(x).

Use the formula$g\left(\alpha \right)=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)e^{-i\alpha x}dx$to find the Fourier transform.

$\begin{array}{c}g\left(\alpha \right)=\frac{1}{2\pi }\underset{-\pi /2}{\overset{\pi /2}{\int }}\frac{1}{2}(e^{ix}+e^{-ix})e^{-i\alpha x}dx\\ =\frac{1}{4\pi }\underset{-\pi /2}{\overset{\pi /2}{\int }}(e^{ix(1-\alpha )}+e^{-ix(1+\alpha )})dx\\ ={\frac{1}{4\pi }[\frac{1}{i(1-\alpha )}{e}^{ix(1-\alpha )}-\frac{1}{i(1+\alpha )}{e}^{-ix(1+\alpha )}]|}_{-\pi /2}^{\pi /2}\\ =\frac{1}{2\pi }\frac{\sin (\pi (1-\alpha )/2)}{1-\alpha }+\frac{1}{2\pi }\frac{\sin (\pi (1+\alpha )/2)}{1+\alpha }\end{array}$

Solve further to obtain$g\left(\alpha \right)$.

$\begin{array}{c}g\left(\alpha \right)=\frac{1}{2\pi }[\frac{\cos (\alpha \pi /2)}{1-\alpha }+\frac{\cos (\alpha \pi /2)}{1+\alpha }]\\ =\frac{1}{2\pi }\frac{\cos (\alpha \pi /2)(1-\alpha )+\cos (\alpha \pi /2)(1+\alpha )}{1-{\alpha }^2}\\ =\frac{1}{\pi }\frac{\cos (\alpha \pi /2)}{1-{\alpha }^2}\end{array}$

Now, write f(x) as a Fourier integral using the formula $f\left(x\right)=\underset{-\infty }{\overset{\infty }{\int }}g\left(\alpha \right)e^{i\alpha x}d\alpha$

$f\left(x\right)=\frac{1}{\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{\text{cos}(\alpha \pi /2)}{1-{\alpha }^2}{e}^{i\alpha x}d\alpha$