Fourier transform of $f (x)=(1-|x|)\,\mathbf 1_{|x| < 1}$ I want to find the Fourier transform of $f(x)
 = 
 \left\{\begin{array}{cc}
 1-|x|, & |x| < 1 \\
 0, & |x| > 1 \\
 \end{array}\right.
 $
I found the question already posted here, but no answers were given and the solution OP posted is one I don't entirely agree with.
I calculate $$\hat{f}(t)=\int_{-1}^1 (1-|x|)e^{-iwx}dx=\int_{-1}^0 (1-x)e^{-iwx}dx + \int_{0}^1 (1+x)e^{-iwx}dx$$
$$= \int_{-1}^1 e^{-iwx}dx - \int_{-1}^0 xe^{-iwx}dx  + \int_{0}^1 xe^{-iwx}dx$$
Now calculating each integral:
(1) $\int_{-1}^1 e^{-iwx}dx = -\frac{1}{iw}e^{-iwx}|_{-1}^1=\frac{1}{iw}(e^{iw}-e^{-iw})$
(2) The second follows by parts $\int_{-1}^0 xe^{-iwx}dx = x\frac{-1}{iw}e^{-iwx}|_{-1}^0+\frac{1}{iw}\int_{-1}^0 e^{-iwx}=(\frac{-x}{iw}-\frac{-x}{iw}e^{iw})-\frac{1}{i^2 w^2}e^{-iwx}|_{-1}^0=-\frac{x}{iw}(e^{iw}-1)+\frac{1}{w^2}(1-e^{iw}) = (e^{iw}-1)(\frac{1}{w^2}+\frac{x}{iw})$
(3) The third integral follows using the same solution by parts, with different integration limits $x\frac{-1}{iw}e^{-iwx}|_{0}^1-\frac{1}{i^2 w^2}e^{-iwx}|_{0}^1=\frac{-x}{iw}(e^{-iw}-1)+\frac{1}{w^2}(e^{iw}-1) = (e^{iw}-1)(\frac{1}{w^2}-\frac{x}{iw})$.
The Fourier transform then will be given by 
$$\hat{f}(t) = \frac{1}{iw}(e^{iw}-e^{-iw}) + (e^{iw}-1)(\frac{1}{w^2}+\frac{x}{iw}) + (e^{iw}-1)(\frac{1}{w^2}-\frac{x}{iw})\\= \frac{\sin(w)}{w} + \frac{2}{w^2}(e^{iw}-1)$$
 A: There was an error in the OP regarding the calculation of $1-|x|$.  
Instead note that we have for $-1\le x\le 1$
$$1-|x|=\begin{cases}1+x&,-1\le x<0\\\\1-x&,0\le x\le 1\end{cases}$$
And inasmuch at $1-|x|$ is an even function about $x=0$, we can write
$$\hat f(t)=2\int_0^1 (1-x)\cos(\omega x)\,dx=\left(\frac{2\sin(\omega/2)}{\omega}\right)^2$$
where we used $1-\cos(\omega)=2\sin^2(\omega/2)$.
A: Method 1
Let be $$\Pi(x)=\begin{cases}1& |x|\le\frac12\\
0&\text{otherwise}\end{cases}$$
the unit box function with
$$
\mathcal F\{\Pi(x)\}=\frac{\sin(\omega/2)}{\omega/2}=\mathrm{Sinc}(\omega/2)
$$
The function 
$$
f(x)=\begin{cases}
 1-|x|, & |x| < 1 \\
 0, & |x| > 1 \\
 \end{cases}
$$
ca be written as $f(x)=(\Pi*\Pi)(x)$ and then
$$
\mathcal F\{f(x)\}=\left[\mathcal F\{\Pi(x)\}\right]^2=\left[\frac{\sin(\omega/2)}{\omega/2}\right]^2=\mathrm{Sinc}^2(\omega/2)
$$
Method 2
$$
\begin{align}
F(\omega)&=\int_{-\infty}^{\infty}f(x)\mathrm e^{-i\omega x}\mathrm d x\\
&=\int_{-1}^0(1+x)\mathrm e^{-i\omega x}\mathrm d x+\int_{0}^1(1-x)\mathrm e^{-i\omega x}\mathrm d x\\
&=\left[\frac{i\omega+1}{\omega^2}-\frac{\mathrm e^{i\omega}}{\omega^2}\right]-\left[\frac{i\omega-1}{\omega^2}-\frac{\mathrm e^{-i\omega}}{\omega^2}\right]\\
&=-\frac{\mathrm e^{-i\omega}\left(\mathrm e^{i\omega}-1\right)^2}{\omega^2}\\
&=-\frac{(2i)^2\sin^2(\omega/2)}{\omega^2}\\
&=\left[\frac{\sin(\omega/2)}{\omega/2}\right]^2\\
&=\mathrm{Sinc}^2(\omega/2)
\end{align}
$$
