Let $ f(x)= \begin{cases} \frac{1}{2}, \mid x \mid<\frac{1}{2} \\ 0, \mid x \mid\geq \frac{1}{2} \end{cases} $
Find Fourier transform
Find the Fourier transform of the convolution $*_{i=1}^{n}f(x)=f(x)*f(x)*\cdots*f(x)$
The function is even as
if $x=\frac{1}{3}$ we get $f(-x)=f(-\frac{1}{3}))=\frac{1}{2}=f(\frac{1}{3})=f(x)$
if $x=5$ we get $f(x-)=f(-5)=0==f(5)=f(x)$
So we have to evaluate $$\hat{f(x)}=\frac{1}{\pi}\int_{0}^{\infty}\frac{1}{2}\cos(\omega x)dx=\frac{1}{2\pi}\int_0^{\frac{1}{2}}\cos(\omega x)dx$$
If $\omega=0$ we have: $$\frac{1}{2\pi}\int_0^{\frac{1}{2}}dx=\frac{1}{4 \pi}$$
If $\omega\neq 0$ we have:
$$\frac{1}{2\pi}\int_0^{\frac{1}{2}}\cos(\omega x)dx=\frac{1}{2\pi}\left.\frac{\sin(\omega x)}{\omega}\right|_0^{\frac{1}{2}}=\frac{\sin(\frac{\omega}{2})}{{2\pi}\omega}$$
We just take $$2\pi \hat{f}(x)^n= \frac{2\pi}{(4 \pi)^n}$$ for $\omega=0$
and
$$2\pi \hat{f}(x)^n=\frac{\sin^n(\frac{\omega}{2})}{\omega^n}$$
For $\omega\neq 0$
Is it correct? we change the limit of integration from $\infty$ to $\frac{1}{2}$ because it is zero whenever $|x|\geq \frac{1}{2}$?