Find the exponential Fourier transform of

$\mathbf{\text{f}}\left(\text{x}\right)\mathbf{\text{=}}\{\begin{array}{l}2a-\left|x\right|,\left|x\right|<2a\\ 0,\left|x\right|>2a\end{array}$

And use your result with Parseval’s theorem to Evaluate${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{4}}\left(\alpha a\right)}{{\mathbf{\alpha }}^{\mathbf{4}}}\mathit{d}\mathit{\alpha }$

The given function is$\text{f}\left(\text{x}\right)\text{=}\left\{\begin{array}{l}2a-\left|x\right|,\left|x\right|<2a\\ 0,\left|x\right|>2a\end{array}\right.$

For the given function u(x) define its Fourier transformation as

$F\left\{\text{u}\left(\text{x}\right)\right\}=\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{-\infty }^{\infty }{\text{e}}^{\text{- ikx}}\text{u}\left(\text{x}\right)\text{dx}$

The Fourier transformation of the function is equal to

$$\begin{gathered} g\left(\alpha \right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{-2a}^0\left(\text{2a + x}\right)e^{-i\alpha x}dx+\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{-2a}^{}\left(\text{2a - x}\right)e^{-i\alpha x}dx \\ =\sqrt{\frac{2}{\mathrm{\pi }}}{\int }_0^{2a}\left(2a-x\right)\cos \left(\alpha x\right)dx \\ =2a\sqrt{\frac{2}{\mathrm{\pi }}}{\int }_0^{2a}\cos \left(\alpha x\right)dx-\sqrt{\frac{2}{\mathrm{\pi }}}{\int }_0^{2a}x\cos \left(\alpha x\right)dx \\ =2a\sqrt{\frac{2}{\mathrm{\pi }}}{\left[\frac{\sin \left(\alpha x\right)}{\alpha }\right]}_0^{2a}-\sqrt{\frac{2}{\mathrm{\pi }}}{\left[\frac{x}{a}\sin \left(\alpha x\right)+\frac{1}{a^2}\cos \left(\alpha x\right)\right]}_0^{2a} \end{gathered}$$

Solve further,

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

Where in the second line we used the fact that the function is even, so it is equal to its Fourier cosine transformation

The integral of${\text{f}}^{\text{2}}\left(\text{x}\right)$

$$\begin{gathered} \underset{\text{-}\infty }{\overset{\infty }{\int }}{\text{f}}^{\text{2}}\left(\text{x}\right)\text{dx = 2}\underset{\text{0}}{\overset{\text{2a}}{\int }}{\left(\text{2a - x}\right)}^{\text{2}}\text{d}\left(\text{2a - x}\right)\text{dx}\left(\text{- 1}\right) \\ \text{=}{\left[\text{-}\frac{\text{2}}{\text{3}}{\left(\text{2a - x}\right)}^{\text{3}}\right]}_{\text{0}}^{\text{2a}} \\ \text{=}\frac{{\text{16a}}^{\text{3}}}{\text{3}} \end{gathered}$$

Using the Parseval theorem

$$\begin{gathered} {\int }_{-\infty }^{\infty }{\left|f\left(x\right)\right|}^{2}dx=\frac{16}{3}{a}^3={\int }_{-\infty }^{\infty }{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{16}{\mathrm{\pi }}{\int }_0^{\infty }\frac{{\sin }^4\left(\alpha \right)}{{\alpha }^4}d\alpha \\ {\int }_0^{\infty }\frac{{\sin }^4\left(\alpha a\right)}{{\alpha }^4}d\alpha =\frac{{\mathrm{\pi a}}^3}{3} \end{gathered}$$

Therefore, the Fourier transformation of the function is equal to $2\sqrt{\frac{2}{\pi }}\frac{{\sin }^2\left(a\alpha \right)}{{\alpha }^2}$and${\int }_0^{\infty }\frac{{\sin }^4\left(\alpha a\right)}{{\alpha }^4}d\alpha =\frac{\pi a^3}{3}$