$\newcommand{\rect}{\operatorname{rect}}$Given $f(x) = \cos(x)\rect(x^2-1) $ , I have to calculate the Fourier transform.
$$F(p) = \frac{1}{\sqrt2\pi} \int_{-\infty} ^\infty \cos(x)\rect(x^2-1) e^{-ipx} \, dx $$
Now, those should be two rects centered in $-1$ and $1$, right?
Should I integrate considering every rect's span $1 $ or
$-\sqrt\frac{3}{2}$, $-\frac{1}{\sqrt2}$ and $+\frac{1}{\sqrt2}$ , $+\sqrt\frac{3}{2}$?
where $$ \rect(t) = \begin{cases} 1 & \quad \text{if } t \in [-\frac{1}{2}, \frac{1}{2}] \\ 0 & \quad \text{otherwise} \end{cases} $$
Edit:
So it should be
$F(p) = \frac{1}{2\sqrt2\pi} \int_{-\sqrt\frac{3}{2}} ^{-\frac{1}{\sqrt2}} e^{ix(1-p)} + e^{-ix(1+p)}dx + \int_{\frac{1}{\sqrt2}} ^{\sqrt\frac{3}{2}} e^{ix(1-p)} + e^{-ix(1+p)}dx $
right?
Edit II:
$\frac{e^{-i\frac{1}{\sqrt2}(1-p)}- e^{-i\frac{\sqrt3}{\sqrt2}(1-p)}}{i(1-p)} + \frac{e^{i\frac{1}{\sqrt2}(1+p)}- e^{i\frac{\sqrt3}{\sqrt2}(1+p)}}{i(1+p)} + \frac{e^{i\frac{\sqrt3}{\sqrt2}(1-p)}- e^{i\frac{1}{\sqrt2}(1-p)}}{-i(1-p)} +\frac{e^{-i\frac{\sqrt3}{\sqrt2}(1+p)}- e^{-i\frac{1}{\sqrt2}(1+p)}}{-i(1+p)}$
So I get this, that I should be able to group up in a sinc, maybe?