Question

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${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{s}\mathbf{i}{\mathbf{n}}^{\mathbf{4}}\left(\alpha a\right)}{{\mathbf{\alpha }}^{\mathbf{4}}}\mathit{d}\mathit{\alpha }$

Short answer

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

Step-by-step solution

Given information

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

Definition of the Fourier transform

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

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

Evaluate Fourier transform of the given function

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

Find the integral of square of the given function

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${\int }_0^{\infty }\frac{{\sin }^4\left(\alpha a\right)}{{\alpha }^4}d\alpha =\frac{\pi a^3}{3}$