91Ó°ÊÓ

The for Fourier series formula gives an expansion of a periodic function f (x) in terms of an horizonless sum of sines and cosines. It's used to putrefy any periodic function or periodic signal into the sum of a group of straightforward oscillating functions, videlicet sines and cosines.

The given functions are

$f\left(x\right)=\{\begin{array}{l}\begin{array}{ll}1,& 0<\mathrm{x}<\frac{\mathrm{Ï€}}{2}\end{array}\\ \begin{array}{ll}0,& \mathrm{x}>\frac{\mathrm{Ï€}}{2}\end{array}\end{array}$

There need to represent the given function as Fourier cosine integral and Fourier sine integral.

Use the Fourier cosine transform formula

$f_c\left(x\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}{∫}_0^{\mathrm{∞}}{g}_c\left(x\right)\cos \mathrm{Î\pm }xd\mathrm{Î\pm }$

Firstly, find the Fourier cosine transform $g_c\left(\mathrm{Î\pm }\right)$is

$\begin{array}{c}{g}_c\left(\mathrm{Î\pm }\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}{∫}_0^{\mathrm{Ï€}/2}\cos \left(\mathrm{Î\pm }x\right)dx\\ ={\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{\sin \left(\mathrm{Î\pm }x\right)}{\mathrm{Î\pm }}|}_0^{\mathrm{Ï€}/2}\\ =\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{\sin \left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)}{\mathrm{Î\pm }}\end{array}$

Substitute the $g_c\left(\mathrm{Î\pm }\right)$in the Fourier cosine transform formula

$\begin{array}{c}{f}_c\left(x\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}{∫}_0^{\mathrm{∞}}\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{\sin \left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)}{\mathrm{Î\pm }}\cos \mathrm{Î\pm }xd\mathrm{Î\pm }\\ f_c\left(x\right)=\frac{2}{\mathrm{Ï€}}{∫}_0^{\mathrm{∞}}\frac{\sin \left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)}{\mathrm{Î\pm }}\cos \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }\end{array}$

Use the Fourier sine transform formula

$f_s\left(x\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}{∫}_0^{\mathrm{∞}}{g}_s\left(\mathrm{Î\pm }\right)\sin \mathrm{Î\pm }xd\mathrm{Î\pm }$

Firstly, find the Fourier sine transform $g_s\left(\mathrm{Î\pm }\right)$ is

$\begin{array}{c}{g}_s\left(\mathrm{Î\pm }\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}{∫}_0^{\mathrm{Ï€}/2}\sin \left(\mathrm{Î\pm }x\right)dx\\ ={−\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{\cos \left(\mathrm{Î\pm }x\right)}{\mathrm{Î\pm }}|}_0^{\mathrm{Ï€}/2}\\ =\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{1−\cos \left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)}{\mathrm{Î\pm }}\end{array}$

Substitute the $g_s\left(\mathrm{Î\pm }\right)$in the Fourier sine transform formula

$\begin{array}{c}{f}_s\left(x\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}{∫}_0^{\mathrm{∞}}\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{1−\cos \left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)}{\mathrm{Î\pm }}\sin \mathrm{Î\pm }xd\mathrm{Î\pm }\\ f_s\left(x\right)=\frac{2}{\mathrm{Ï€}}{∫}_0^{\mathrm{∞}}\frac{1−\cos \left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)}{\mathrm{Î\pm }}\sin \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }\end{array}$

Thus, Fourier sine integral is $f_s\left(x\right)=\frac{2}{\mathrm{Ï€}}{∫}_0^{\mathrm{∞}}\frac{1−\cos \left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)}{\mathrm{Î\pm }}\sin \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }$and the Fourier cosine integral is$f_c\left(x\right)=\frac{2}{\mathrm{Ï€}}{∫}_0^{\mathrm{∞}}\frac{\sin \left(\frac{\mathrm{Î\pm }\mathrm{Ï€}}{2}\right)}{\mathrm{Î\pm }}\cos \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }$.