= Solution
The <triangular function> is real and even, so its <Fourier transform> reduces to a cosine integral:
$$
\widetilde f(k)=2\int_0^1(1-x)\cos(kx)dx.
$$
For $k\ne0$, one <integration by parts> gives $\int_0^1(1-x)\cos(kx)dx=(1-\cos k)/k^2$. Therefore the <Fourier transform of a triangular function> is
$$
\boxed{\widetilde f(k)=\frac{2(1-\cos k)}{k^2}=\left(\frac{\sin(k/2)}{k/2}\right)^2,\qquad\widetilde f(0)=1.}
$$
The zero-frequency value is the removable limit and also the area of the triangle. The transform is a <squared sinc function>, which explains its nonnegativity despite the oscillating Fourier exponential.
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