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
$$\hat{f}(t) = \frac{1}{iw}(e^{iw}-e^{-iw}) + (e^{iw}-1)(\frac{1}{w^2}+\frac{x}{iw}) + (e^{iw}-1)(\frac{1}{w^2}-\frac{x}{iw})\\= \frac{\sin(w)}{w} + \frac{2}{w^2}(e^{iw}-1)$$
Best Answer
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
$$ \begin{align} F(\omega)&=\int_{-\infty}^{\infty}f(x)\mathrm e^{-i\omega x}\mathrm d x\\ &=\int_{-1}^0(1+x)\mathrm e^{-i\omega x}\mathrm d x+\int_{0}^1(1-x)\mathrm e^{-i\omega x}\mathrm d x\\ &=\left[\frac{i\omega+1}{\omega^2}-\frac{\mathrm e^{i\omega}}{\omega^2}\right]-\left[\frac{i\omega-1}{\omega^2}-\frac{\mathrm e^{-i\omega}}{\omega^2}\right]\\ &=-\frac{\mathrm e^{-i\omega}\left(\mathrm e^{i\omega}-1\right)^2}{\omega^2}\\ &=-\frac{(2i)^2\sin^2(\omega/2)}{\omega^2}\\ &=\left[\frac{\sin(\omega/2)}{\omega/2}\right]^2\\ &=\mathrm{Sinc}^2(\omega/2) \end{align} $$