Represent each of the following functions (a) by a Fourier cosine integral; (b) by a Fourier sine integral. Hint: See the discussion just before Parseval’s theorem.

29.$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}-1,0<x<2\\ 1,2<x<3\\ 0,x>3\end{array}$

The given function is$f\left(x\right)=\left\{\begin{array}{l}-1,0<x<2\\ 1,2<x<3\\ 0,x>3\end{array}\right.$.Fourier cosine integral and Fourier sine integral is to be found out of this function.

The Fourier integral theorem says that, if a function $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$satisfies the Dirichlet conditions on every finite interval and if localid="1664273428560" $\underset{\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\mathbf{\left|}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right|}\mathit{d}\mathit{x}$is finite, then localid="1664273443579" role="math" $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{-}\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\mathit{g}\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}{\mathit{e}}^{\mathbf{i}\mathbf{\alpha }\mathbf{x}}\mathit{d}\mathit{\alpha }$andlocalid="1664273469575" $\mathit{g}\mathbf{\left(}\mathit{\alpha }\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}\underset{\mathbf{-}\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{\alpha }\mathbf{x}}\mathit{d}\mathit{x}$.

$\begin{array}{c}{g}_c\left(\alpha \right)=\sqrt{\frac{2}{\pi }}[-\underset{0}{\overset{2}{\int }}\cos \left(\alpha x\right)dx+\underset{2}{\overset{3}{\int }}\cos \left(\alpha x\right)dx]\\ =\sqrt{\frac{2}{\pi }}[{(-\frac{\sin \left(\alpha x\right)}{\alpha })}_0^2+{\left(\frac{\sin \left(\alpha x\right)}{\alpha }\right)}_2^3]\\ =\sqrt{\frac{2}{\pi }}\left[\frac{\sin \left(3\alpha \right)-2\sin \left(2\alpha \right)}{\alpha }\right]\end{array}$

Thus, $f_c\left(x\right)=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\sin \left(3\alpha \right)-2\sin \left(2\alpha \right)}{\alpha }\cos \left(\alpha x\right)d\alpha$.

$\begin{array}{c}{g}_s\left(\alpha \right)=\sqrt{\frac{2}{\pi }}[-\underset{0}{\overset{2}{\int }}\sin \left(\alpha x\right)dx+\underset{2}{\overset{3}{\int }}\sin \left(\alpha x\right)dx]\\ =\sqrt{\frac{2}{\pi }}[{\left(\frac{\cos \left(\alpha x\right)}{\alpha }\right)}_0^2-{\left(\frac{\cos \left(\alpha x\right)}{\alpha }\right)}_2^3]\\ =\sqrt{\frac{2}{\pi }}\left[\frac{\cos \left(2\alpha \right)-1-\cos \left(3\alpha \right)}{\alpha }\right]\end{array}$

Thus, $f_s\left(x\right)=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{2\cos \left(2\alpha \right)-1-\cos \left(3\alpha \right)}{\alpha }\sin \left(\alpha x\right)d\alpha$

Therefore, Fourier cosine integral is $f_c\left(x\right)=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\sin \left(3\alpha \right)-2\sin \left(2\alpha \right)}{\alpha }\cos \left(\alpha x\right)d\alpha$And Fourier sine integral is $f_s\left(x\right)=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{2\cos \left(2\alpha \right)-1-\cos \left(3\alpha \right)}{\alpha }\sin \left(\alpha x\right)d\alpha$.