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

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

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

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

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

Use the Fourier cosine transform formula

$f_c\left(x\right)=\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }{g}_c\left(x\right)\cos \alpha xd\alpha$

Firstly, find the Fourier cosine transform $g_c\left(\alpha \right)$is

$\begin{array}{c}{g}_c\left(\alpha \right)=\sqrt{\frac{2}{\pi }}{\int }_0^{\pi /2}\cos \left(\alpha x\right)dx\\ ={\sqrt{\frac{2}{\pi }}\frac{\sin \left(\alpha x\right)}{\alpha }|}_0^{\pi /2}\\ =\sqrt{\frac{2}{\pi }}\frac{\sin \left(\frac{\alpha \pi }{2}\right)}{\alpha }\end{array}$

Substitute the $g_c\left(\alpha \right)$in the Fourier cosine transform formula

$\begin{array}{c}{f}_c\left(x\right)=\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }\sqrt{\frac{2}{\pi }}\frac{\sin \left(\frac{\alpha \pi }{2}\right)}{\alpha }\cos \alpha xd\alpha \\ f_c\left(x\right)=\frac{2}{\pi }{\int }_0^{\infty }\frac{\sin \left(\frac{\alpha \pi }{2}\right)}{\alpha }\cos \left(\alpha x\right)d\alpha \end{array}$

Use the Fourier sine transform formula

$f_s\left(x\right)=\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }{g}_s\left(\alpha \right)\sin \alpha xd\alpha$

Firstly, find the Fourier sine transform $g_s\left(\alpha \right)$ is

$\begin{array}{c}{g}_s\left(\alpha \right)=\sqrt{\frac{2}{\pi }}{\int }_0^{\pi /2}\sin \left(\alpha x\right)dx\\ ={-\sqrt{\frac{2}{\pi }}\frac{\cos \left(\alpha x\right)}{\alpha }|}_0^{\pi /2}\\ =\sqrt{\frac{2}{\pi }}\frac{1-\cos \left(\frac{\alpha \pi }{2}\right)}{\alpha }\end{array}$

Substitute the $g_s\left(\alpha \right)$in the Fourier sine transform formula

$\begin{array}{c}{f}_s\left(x\right)=\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }\sqrt{\frac{2}{\pi }}\frac{1-\cos \left(\frac{\alpha \pi }{2}\right)}{\alpha }\sin \alpha xd\alpha \\ f_s\left(x\right)=\frac{2}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(\frac{\alpha \pi }{2}\right)}{\alpha }\sin \left(\alpha x\right)d\alpha \end{array}$

Thus, Fourier sine integral is $f_s\left(x\right)=\frac{2}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(\frac{\alpha \pi }{2}\right)}{\alpha }\sin \left(\alpha x\right)d\alpha$and the Fourier cosine integral is$f_c\left(x\right)=\frac{2}{\pi }{\int }_0^{\infty }\frac{\sin \left(\frac{\alpha \pi }{2}\right)}{\alpha }\cos \left(\alpha x\right)d\alpha$.