TT2 problem 5

[Victor Ivrii]

Find Fourier transforms of the  function
\begin{equation*}
f(x)= \left\{\begin{aligned}
&1-x^2 &&|x|<1\\
&0 &&|x|>1
\end{aligned}\right.
\end{equation*}
and write this function as a Fourier integral.

[Mark Nunez]

 
Edited.

[Yiyun Liu]

\[\begin{array}{l}
part(1):\\
\\
\hat f(\omega ) = \frac{1}{{2\pi }}\int\limits_{ - 1}^1 {(1 - {x^2})} {e^{ - i\omega x}}dx\\
 = \frac{1}{{2\pi }}[\int\limits_{ - 1}^1 {{e^{ - i\omega x}}dx - } \int\limits_{ - 1}^1 {{x^2}} {e^{ - i\omega x}}dx]\\
 = \frac{1}{{2\pi }}(\frac{{ - 1}}{{i\omega }}{e^{ - i\omega x}}\mathop |\nolimits_{ - 1}^1 ) - ( - \frac{1}{{i\omega }}{x^2}{e^{ - i\omega x}}\mathop |\nolimits_{ - 1}^1  + \int\limits_{ - 1}^1 {\frac{{2x}}{{i\omega }}} {e^{ - i\omega x}}dx)\\
 =  - \frac{1}{{2\pi }}\int\limits_{ - 1}^1 {\frac{{2x}}{{i\omega }}} {e^{ - i\omega x}}dx\\
 =  - (\frac{2}{{\pi {\omega ^2}}}\cos \omega  - \frac{2}{{\pi {\omega ^3}}}\sin (\omega ))\\
 = \frac{2}{{\pi {\omega ^3}}}\sin (\omega ) - \frac{2}{{\pi {\omega ^2}}}\cos (\omega )\\
\\
part(2):\\
\\
f(x) = \int\limits_{ - 1}^1 {[\frac{2}{{\pi {\omega ^3}}}\sin (\omega ) - \frac{2}{{\pi {\omega ^2}}}\cos (\omega )]} {e^{i\omega x}}d\omega
\end{array}\]

[Victor Ivrii]

Yiyun is right. The simplest way:
$$\hat{f}(k)=\frac{1}{2\pi}\int _{-1}^1 (1-x^2)e^{-ikx}\,dx =
\frac{1}{2\pi} \int _0^1 (1-x^2)\bigl(e^{-ikx}+ e^{ikx}\bigr)\,dx=
\frac{1}{\pi} \int _0^1 (1-x^2)\cos (kx)\,dx$$
where we used that the function is even. Integrating by parts
$$\hat{f}(k)=  \frac{1}{\pi k} (1-x^2)\sin (kx)\bigr|_0^{1} +
\frac{2}{\pi k} \int _0^1 x\sin  (kx)\,dx$$
and the first term is $0$; integrating by parts again
$$\hat{f}(k)=   
\bigl[-\frac{2}{\pi k^2} \cos   (kx) + \frac{2}{\pi k^3} \sin (kx)\bigr]\bigr|_0^{1}=
-\frac{2}{\pi k^2} x\cos   (k) + \frac{2}{\pi k^3} \sin (k)
$$