91Ó°ÊÓ

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" $\underset{\mathbf{∞}}{\overset{\mathbf{∞}}{\mathbf{∫}}}\mathbf{\left|}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right|}\mathit{d}\mathit{x}$is finite, then localid="1664273443579" role="math" $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{−}\mathbf{∞}}{\overset{\mathbf{∞}}{\mathbf{∫}}}\mathit{g}\mathbf{\left(}\mathbf{Î\pm }\mathbf{\right)}{\mathit{e}}^{\mathbf{i}\mathbf{Î\pm }\mathbf{x}}\mathit{d}\mathit{Î\pm }$andlocalid="1664273469575" $\mathit{g}\mathbf{\left(}\mathit{Î\pm }\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{Ï€}}\underset{\mathbf{-}\mathbf{∞}}{\overset{\mathbf{∞}}{\mathbf{∫}}}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{Î\pm }\mathbf{x}}\mathit{d}\mathit{x}$.

$\begin{array}{c}{g}_c\left(\mathrm{Î\pm }\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}[-\underset{0}{\overset{2}{∫}}\cos \left(\mathrm{Î\pm }x\right)dx+\underset{2}{\overset{3}{∫}}\cos \left(\mathrm{Î\pm }x\right)dx]\\ =\sqrt{\frac{2}{\mathrm{Ï€}}}[{(-\frac{\sin \left(\mathrm{Î\pm }x\right)}{\mathrm{Î\pm }})}_0^2+{\left(\frac{\sin \left(\mathrm{Î\pm }x\right)}{\mathrm{Î\pm }}\right)}_2^3]\\ =\sqrt{\frac{2}{\mathrm{Ï€}}}\left[\frac{\sin \left(3\mathrm{Î\pm }\right)-2\sin \left(2\mathrm{Î\pm }\right)}{\mathrm{Î\pm }}\right]\end{array}$

Thus, $f_c\left(x\right)=\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{\sin \left(3\mathrm{Î\pm }\right)-2\sin \left(2\mathrm{Î\pm }\right)}{\mathrm{Î\pm }}\cos \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }$.

$\begin{array}{c}{g}_s\left(\mathrm{Î\pm }\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}[-\underset{0}{\overset{2}{∫}}\sin \left(\mathrm{Î\pm }x\right)dx+\underset{2}{\overset{3}{∫}}\sin \left(\mathrm{Î\pm }x\right)dx]\\ =\sqrt{\frac{2}{\mathrm{Ï€}}}[{\left(\frac{\cos \left(\mathrm{Î\pm }x\right)}{\mathrm{Î\pm }}\right)}_0^2-{\left(\frac{\cos \left(\mathrm{Î\pm }x\right)}{\mathrm{Î\pm }}\right)}_2^3]\\ =\sqrt{\frac{2}{\mathrm{Ï€}}}\left[\frac{\cos \left(2\mathrm{Î\pm }\right)-1-\cos \left(3\mathrm{Î\pm }\right)}{\mathrm{Î\pm }}\right]\end{array}$

Thus, $f_s\left(x\right)=\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{2\cos \left(2\mathrm{Î\pm }\right)-1-\cos \left(3\mathrm{Î\pm }\right)}{\mathrm{Î\pm }}\sin \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }$

Therefore, Fourier cosine integral is $f_c\left(x\right)=\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{\sin \left(3\mathrm{Î\pm }\right)-2\sin \left(2\mathrm{Î\pm }\right)}{\mathrm{Î\pm }}\cos \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }$And Fourier sine integral is $f_s\left(x\right)=\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{2\cos \left(2\mathrm{Î\pm }\right)-1-\cos \left(3\mathrm{Î\pm }\right)}{\mathrm{Î\pm }}\sin \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }$.