91Ó°ÊÓ

The graph of the given function is as shown below.

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

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

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,

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

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.

$\begin{array}{c}g\left(\mathrm{Î\pm }\right)=\frac{1}{2\mathrm{Ï€}}\underset{−a}{\overset{0}{∫}}2(a+x)e^{−i\mathrm{Î\pm }x}dx+\frac{1}{2\mathrm{Ï€}}\underset{0}{\overset{a}{∫}}2(a−x)e^{−i\mathrm{Î\pm }x}dx\\ =\frac{a}{\mathrm{Ï€}}\underset{−a}{\overset{0}{∫}}{e}^{−i\mathrm{Î\pm }x}dx+\frac{a}{\mathrm{Ï€}}\underset{0}{\overset{a}{∫}}{e}^{−i\mathrm{Î\pm }x}dx+\frac{1}{\mathrm{Ï€}}\underset{−a}{\overset{0}{∫}}x{e}^{−i\mathrm{Î\pm }x}dx−\frac{1}{\mathrm{Ï€}}\underset{0}{\overset{a}{∫}}x{e}^{−i\mathrm{Î\pm }x}dx\\ =−\frac{a}{\mathrm{Ï€}}\underset{a}{\overset{0}{∫}}{e}^{i\mathrm{Î\pm }x}dx+\frac{a}{\mathrm{Ï€}}\underset{0}{\overset{a}{∫}}{e}^{−i\mathrm{Î\pm }x}dx+\frac{1}{\mathrm{Ï€}}\underset{a}{\overset{0}{∫}}(−x)e^{i\mathrm{Î\pm }x}d(−x)−\frac{1}{\mathrm{Ï€}}\underset{0}{\overset{a}{∫}}x{e}^{−i\mathrm{Î\pm }x}dx\\ =\frac{2a}{\mathrm{Ï€}}\underset{0}{\overset{a}{∫}}cos\left(\mathrm{Î\pm }x\right)dx−\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{a}{∫}}x\cos \left(\mathrm{Î\pm }x\right)dx\end{array}$

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

$\begin{array}{c}g\left(\mathrm{Î\pm }\right)={\frac{2a}{\mathrm{Î\pm }\mathrm{Ï€}}\text{sin}\left(\mathrm{Î\pm }x\right)|}_0^a−{\frac{2}{\mathrm{Ï€}}[\frac{x}{\mathrm{Î\pm }}\text{sin}\left(\mathrm{Î\pm }x\right)+\frac{1}{{\mathrm{Î\pm }}^2}\text{cos}\left(\mathrm{Î\pm }x\right)]|}_0^a\\ =\frac{2a}{\mathrm{Ï€}\mathrm{Î\pm }}\text{sin}\left(\mathrm{Î\pm }a\right)−\frac{2}{\mathrm{Ï€}}[\frac{a}{\mathrm{Î\pm }}\text{sin}\left(\mathrm{Î\pm }a\right)+\frac{1}{{\mathrm{Î\pm }}^2}(\text{cos}\left(\mathrm{Î\pm }a\right)−1)]\\ =\frac{2}{\mathrm{Ï€}}\frac{1−\text{cos}\left(\mathrm{Î\pm }a\right)}{{\mathrm{Î\pm }}^2}\end{array}$

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

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