Question

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.

28.$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}1,2<x<4\\ 0,0<x<2,x>4\end{array}$

Short answer

(a) Fourier cosine integral is $f_c\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\cos \left(3\alpha \right)\sin \alpha }{\alpha }\cos \left(\alpha x\right)d\alpha$.

(b) Fourier sine integral is $f_s\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\sin \left(3\alpha \right)\sin \alpha }{\alpha }\sin \left(\alpha x\right)d\alpha$.

Step-by-step solution

Given Information.

The given function is $f\left(x\right)=\left\{\begin{array}{l}1,2<x<4\\ 0,0<x<2,x>4\end{array}\right.$.Fourier cosine integral and Fourier sine integral is to be found out of this function.

Definition of Fourier integral theorem

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 $\underset{\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\mathbf{\left|}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right|}\mathit{d}\mathit{x}$is finite, then $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{-}\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\mathit{g}\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}{\mathit{e}}^{\mathbf{i}\mathbf{\alpha }\mathbf{x}}\mathit{d}\mathit{\alpha }$and$\mathit{g}\mathbf{\left(}\mathit{\alpha }\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}\underset{\mathbf{-}\mathbf{\infty }}{\overset{\mathbf{\infty }}{\mathbf{\int }}}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{\alpha }\mathbf{x}}\mathit{d}\mathit{x}$.

Find the cosine integral

$\begin{array}{c}{g}_c\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}f\left(x\right)\cos \left(\alpha x\right)dx\\ =\sqrt{\frac{2}{\pi }}\underset{2}{\overset{4}{\int }}\cos \left(\alpha x\right)dx\\ =\sqrt{\frac{2}{\pi }}{\left[\frac{\sin \left(\alpha x\right)}{\alpha }\right]}_2^4\\ =\sqrt{\frac{2}{\pi }}\frac{1}{\alpha }(\sin \left(4\alpha \right)-\sin \left(2\alpha \right))\end{array}$

Now $g_c\left(\alpha \right)=\frac{2}{\alpha }\sqrt{\frac{2}{\pi }}\cos \left(3\alpha \right)\sin \alpha$

Thus $f_c\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\cos \left(3\alpha \right)\sin \alpha }{\alpha }\cos \left(\alpha x\right)d\alpha$.

Find the sine integral

$\begin{array}{c}{g}_s\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{\infty }{\int }}f\left(x\right)\sin \left(\alpha x\right)dx\\ =\sqrt{\frac{2}{\pi }}\underset{2}{\overset{4}{\int }}\sin \left(\alpha x\right)dx\\ =-\sqrt{\frac{2}{\pi }}{\left[\frac{\cos \left(\alpha x\right)}{\alpha }\right]}_2^4\\ =\sqrt{\frac{2}{\pi }}\frac{1}{\alpha }(\cos \left(2\alpha \right)-\cos \left(4\alpha \right))\end{array}$

Now$g_s\left(\alpha \right)=\frac{2}{\alpha }\sqrt{\frac{2}{\pi }}\sin \left(3\alpha \right)\sin \alpha$

Thus$f_s\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\sin \left(3\alpha \right)\sin \alpha }{\alpha }\sin \left(\alpha x\right)d\alpha$

Therefore, Fourier cosine integral is $f_c\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\cos \left(3\alpha \right)\sin \alpha }{\alpha }\cos \left(\alpha x\right)d\alpha$

And Fourier sine integral is $f_s\left(x\right)=\frac{4}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\sin \left(3\alpha \right)\sin \alpha }{\alpha }\sin \left(\alpha x\right)d\alpha$.