Question

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

$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}\mathrm{sinx},\left|x\right|<\pi /2\\ 0,\left|x\right|>\pi /2\end{array}$

Short answer

The exponential Fourier transform of the given function is $g\left(\alpha \right)=\frac{-i}{\pi }\frac{\alpha }{1-{\alpha }^2}\cos \left(\frac{\alpha \pi }{2}\right)$and f(x) as a Fourier integral is$f\left(x\right)=\frac{-i}{\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{\alpha }{1-{\alpha }^2}\cos \left(\frac{\alpha \pi }{2}\right)e^{i\alpha x}d\alpha$.

Step-by-step solution

Given information

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

Definition of Fourier transforms

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$f\left(x\right)$.

Find the Fourier transform

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.

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

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

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

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

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