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.

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

Short answer

(a) Fourier cosine integral is $f_c\left(x\right)=\frac{1}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{1-\cos \left(2\alpha \right)}{{\alpha }^2}\cos \left(\alpha x\right)d\alpha$.

(b) Fourier sine integral is $f_s\left(x\right)=\frac{1}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{2\alpha -\sin \left(2\alpha \right)}{{\alpha }^2}\sin \left(\alpha x\right)d\alpha$.

Step-by-step solution

Given Information.

The given function is$f\left(x\right)=\{\begin{array}{ll}1-\frac{\mathrm{x}}{2},& 0<\mathrm{x}<2\\ 0,& \mathrm{x}>2\end{array}$.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 $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\underset{-\infty }{\overset{\infty }{\int }}g\left(\alpha \right){\mathit{e}}^{\mathbf{i}\mathbf{\alpha }\mathbf{x}}\mathit{d}\mathit{\alpha }$and$\mathit{g}\mathbf{\left(}\mathit{\alpha }\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right){\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{\alpha }\mathbf{x}}\mathit{d}\mathit{x}$.

Find the cosine integral

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

Thus localid="1664272783731" $f_c\left(x\right)=\frac{1}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{1-\cos \left(2\alpha \right)}{{\alpha }^2}\cos \left(\alpha x\right)d\alpha$.

Find the sine integral

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

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

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

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