How does one derive the Fourier transform of the Ramp function? One approach could have been to see that the Ramp function ( http://mathworld.wolfram.com/RampFunction.html ) is the convolution of $2$ Heavisides (at $0$). Hence its Fourier transform should have been the product of the Fourier transforms of Heavisides. The Fourier transform of the Heaviside (http://mathworld.wolfram.com/HeavisideStepFunction.html) is, $\frac{1}{2} [\delta(t) - \frac{1}{\pi t}  ] $. But its not clear to me as to how its square is the Fourier transform of the Ramp at $0$ which is $\frac{i}{4\pi} \delta'(t) - \frac{1}{4\pi^2 t^2} $


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*I would otherwise like to see a reference (or if someone can type in!) which derives the Fourier transform of the ramp function from scratch! 


In fact the convolution theorem is a bit more complicated than you think for distributions. To apply it you need to approximate your distributions with something like functions $\in L^1$. So 
$$1_{ x > 0} = \lim_{a \to 0^+} e^{-ax} 1_{x > 0}$$
$$x 1_{ x > 0} = \lim_{a \to 0^+} e^{-ax} 1_{x > 0} \ast e^{-ax} 1_{x > 0}$$
$$\mathcal{F}[x 1_{ x > 0}] = \lim_{a \to 0^+} \mathcal{F}[e^{-ax} 1_{x > 0} \ast e^{-ax} 1_{x > 0}]$$
$$ = \lim_{a \to 0^+} \mathcal{F}[e^{-ax} 1_{x > 0} ]^2$$
$$ = \lim_{a \to 0^+} \frac{1}{(a+2i \pi \xi)^2}$$
$$ = \lim_{a \to 0^+}  \frac{-1}{4\pi^2} \frac{1}{(\xi-ia)^2}$$
$$ = \lim_{a \to 0^+}  \frac{1}{4\pi^2} \frac{d^2}{d\xi^2} \log(\xi-ia)$$
$$ = \lim_{a \to 0^+}  \frac{1}{4\pi^2}\frac{d^2}{d\xi^2}( -i\pi 1_{\xi 
< 0} +\log|\xi|)$$
$$ = \frac{i}{4\pi} \delta'(\xi)- \frac{1}{4 \pi^2 } fp(\frac{1}{\xi^2})$$
where $fp(\frac{1}{\xi^2})$ is the finite part the second derivative of $-\log |\xi|$
