Question

Given

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\left(\begin{array}{c}x,\\ 2-x,\\ 0\end{array}\begin{array}{ccccc}0& \le & x& \le & 1\\ 1& \le & x& \le & 2\\ & x& \ge & 2& \end{array}\right)$

find the cosine transform of f(x) and f(x) use it to write as an integral. Use your result to evaluate

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{\alpha }\mathbf{}\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{2}}\mathbf{\alpha }\mathbf{/}\mathbf{2}}{{\mathbf{\alpha }}^{\mathbf{2}}}\mathit{d}\mathit{\alpha }$

Short answer

The cosine transformis $f\left(x\right)=\frac{8}{\mathrm{\pi }}{\int }_0^{\infty }\frac{\cos \alpha {\sin }^2\left(\alpha /2\right)}{{\alpha }^2}\cos \left(\alpha x\right)d\alpha$

and the value of the integral is${\int }_0^{\infty }\frac{{\cos }^2\alpha {\sin }^2\left(\alpha /2\right)}{{\alpha }^2}d\alpha =\frac{\mathrm{\pi }}{8}$

Step-by-step solution

Given Information

The given equation is$f\left(x\right)=\left(\begin{array}{c}x,\\ 2-x,\\ 0\end{array}\begin{array}{ccccc}0& \le & x& \le & 1\\ 1& \le & x& \le & 2\\ & x& \ge & 2& \end{array}\right)$

Step 2: Meaning of the Fourier Series.

An infinite sum of sines and cosines is used to represent the expansion of a periodic function f(x) into a Fourier series. The orthogonality relationships between the sine and cosine functions are used in the Fourier series.

Find the cosine transform and evaluate the integral.

The Fourier cosine transform is equal to:

$$\begin{gathered} g_c\alpha =\sqrt{\frac{2}{\mathrm{\pi }}}{\int }_0^{1}x\cos \left(\alpha x\right)dx+\sqrt{\frac{2}{\mathrm{\pi }}}{\int }_1^2\left(2-x\right)\cos \left(\alpha x\right)dx \\ =\sqrt{\frac{2}{\mathrm{\pi }}}\left[\left(\frac{x}{\alpha }\sin \left(\alpha x\right)+\frac{1}{{\alpha }^2}\cos \left(\alpha x\right)\right){\left|+2\frac{\sin \left(\alpha x\right)}{\alpha }\right|}_1^2-{\left(\frac{x}{\alpha }\sin \left(\alpha x\right)+\frac{1}{{\alpha }^2}\cos \left(\alpha x\right)\right)}_1^2\right] \\ =\sqrt{\frac{2}{\mathrm{\pi }}}\frac{1}{{\alpha }^2}\left(\cos \alpha -1-\cos \left(2\alpha \right)+\cos \alpha \right) \\ =\sqrt{\frac{2}{\mathrm{\pi }}}\frac{1}{{\alpha }^2}\left(2\cos \alpha -\cos \left(2\alpha \right)-1\right) \end{gathered}$$

Further solving

$$\begin{gathered} g_c\alpha =\sqrt{\frac{2}{\mathrm{\pi }}}\frac{1}{{\alpha }^2}\left(2\cos \alpha -2{\cos }^2\alpha \right) \\ =\sqrt{\frac{2}{\mathrm{\pi }}}\frac{2}{{\alpha }^2}\cos \alpha \left(1-\cos \alpha \right) \\ =\sqrt{\frac{2}{\mathrm{\pi }}}\frac{4}{{\alpha }^2}\cos \alpha {\sin }^2\frac{\alpha }{2} \end{gathered}$$

Thus, the function is equal to

$f\left(x\right)=\frac{8}{\mathrm{\pi }}{\int }_0^{\infty }\frac{\cos \alpha {\sin }^2\left(\alpha /2\right)}{{\alpha }^2}d\alpha$

Now set x=1 , and f(1)=1 use

$$\begin{gathered} 1=\frac{8}{\mathrm{\pi }}{\int }_0^{\infty }\frac{{\cos }^2\alpha {\sin }^2\left(\alpha /2\right)}{{\alpha }^2}d\alpha \\ {\int }_0^{\infty }\frac{{\cos }^2\alpha {\sin }^2\left(\alpha /2\right)}{{\alpha }^2}d\alpha =\frac{\mathrm{\pi }}{8} \end{gathered}$$

Therefore,the cosine transform is$f\left(x\right)=\frac{2}{\mathrm{\pi }}{\int }_0^{\infty }\frac{\cos \alpha {\sin }^2\left(\alpha /2\right)}{{\alpha }^2}\cos \left(\alpha x\right)d\alpha$

and the value of the integral is

${\int }_0^{\infty }\frac{{\cos }^2\alpha {\sin }^2\left(\alpha /2\right)}{{\alpha }^2}d\alpha =\frac{\mathrm{\pi }}{8}$