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

The graph of the given function is as shown below.

In mathematical form, the function can be written as follows.

$f\left(x\right)=\left\{\begin{array}{l}2(x+a),x\in [-a,0]\\ 2(a-x),x\in [0,a]\end{array}\right.$

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.

$\begin{array}{c}g\left(\alpha \right)=\frac{1}{2\pi }\underset{-a}{\overset{0}{\int }}2(a+x)e^{-i\alpha x}dx+\frac{1}{2\pi }\underset{0}{\overset{a}{\int }}2(a-x)e^{-i\alpha x}dx\\ =\frac{a}{\pi }\underset{-a}{\overset{0}{\int }}{e}^{-i\alpha x}dx+\frac{a}{\pi }\underset{0}{\overset{a}{\int }}{e}^{-i\alpha x}dx+\frac{1}{\pi }\underset{-a}{\overset{0}{\int }}x{e}^{-i\alpha x}dx-\frac{1}{\pi }\underset{0}{\overset{a}{\int }}x{e}^{-i\alpha x}dx\\ =-\frac{a}{\pi }\underset{a}{\overset{0}{\int }}{e}^{i\alpha x}dx+\frac{a}{\pi }\underset{0}{\overset{a}{\int }}{e}^{-i\alpha x}dx+\frac{1}{\pi }\underset{a}{\overset{0}{\int }}(-x)e^{i\alpha x}d(-x)-\frac{1}{\pi }\underset{0}{\overset{a}{\int }}x{e}^{-i\alpha x}dx\\ =\frac{2a}{\pi }\underset{0}{\overset{a}{\int }}cos\left(\alpha x\right)dx-\frac{2}{\pi }\underset{0}{\overset{a}{\int }}x\cos \left(\alpha x\right)dx\end{array}$

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

$\begin{array}{c}g\left(\alpha \right)={\frac{2a}{\alpha \pi }\text{sin}\left(\alpha x\right)|}_0^a-{\frac{2}{\pi }[\frac{x}{\alpha }\text{sin}\left(\alpha x\right)+\frac{1}{{\alpha }^2}\text{cos}\left(\alpha x\right)]|}_0^a\\ =\frac{2a}{\pi \alpha }\text{sin}\left(\alpha a\right)-\frac{2}{\pi }[\frac{a}{\alpha }\text{sin}\left(\alpha a\right)+\frac{1}{{\alpha }^2}(\text{cos}\left(\alpha a\right)-1)]\\ =\frac{2}{\pi }\frac{1-\text{cos}\left(\alpha a\right)}{{\alpha }^2}\end{array}$

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

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