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

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

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

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.

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

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