The signum function is defined by$$sgn(t)=\left\{\begin{matrix}-1, \ t<0\\0, \ t=0 \\ 1, \ t>0 \end{matrix}\right.$$has derivative$$\frac{d}{dt} sign(t) = 2 \delta(t)$$Use this result to show that $j2\pi \nu S(\nu)=2,$ and give an argument why $S(0)=0.$ Where $S(\nu)$ denotes the Fourier transform of the signum function
Attempt
I already know that the Fourier transform of the sign function is:
$$S(\nu)=\frac{1}{j\pi \nu}.$$
Of course, if we substitute this into $j2\pi \nu S(\nu),$ we get $2$ as expected. But the question wants us to show this result without deriving an expression for the Fourier transform of $sgn(t).$ How can we do this?
Furthermore, if we substitute $0$ in $S(\nu)$ we get $S(0)=1/j\pi (0) = \infty.$ So what does the question mean by giving an argument that $S(0)=0$?