91Ó°ÊÓ

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

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

The Fourier transformation of the function is equal to

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

Solve further,

$$\begin{gathered} =\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{1-\cos \left(2a\mathrm{Î\pm }\right)}{{\mathrm{Î\pm }}^2} \\ =2\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{{\sin }^2\left(a\mathrm{Î\pm }\right)}{{\mathrm{Î\pm }}^2} \end{gathered}$$

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

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

Using the Parseval theorem

$$\begin{gathered} {∫}_{-∞}^{∞}{\left|f\left(x\right)\right|}^{2}dx=\frac{16}{3}{a}^3={∫}_{-∞}^{∞}{\left|g\left(\mathrm{Î\pm }\right)\right|}^{2}d\mathrm{Î\pm }=\frac{16}{\mathrm{Ï€}}{∫}_0^{∞}\frac{{\sin }^4\left(\mathrm{Î\pm }\right)}{{\mathrm{Î\pm }}^4}d\mathrm{Î\pm } \\ {∫}_0^{∞}\frac{{\sin }^4\left(\mathrm{Î\pm }a\right)}{{\mathrm{Î\pm }}^4}d\mathrm{Î\pm }=\frac{{\mathrm{Ï€²¹}}^3}{3} \end{gathered}$$

Therefore, the Fourier transformation of the function is equal to $2\sqrt{\frac{2}{\mathrm{Ï€}}}\frac{{\sin }^2\left(a\mathrm{Î\pm }\right)}{{\mathrm{Î\pm }}^2}$and${∫}_0^{∞}\frac{{\sin }^4\left(\mathrm{Î\pm }a\right)}{{\mathrm{Î\pm }}^4}d\mathrm{Î\pm }=\frac{\mathrm{Ï€}{a}^3}{3}$