What is Fourier transform of $|2\sin (2x)|$ I want to know what is the Fourier transform of |2sin(2x)|
I know how to calculate the Fourier transform of 2sin(2x) but absolute value bothers me.
I write it in another form --> $$|e^{-2ix} - e^{2ix} | $$
but it doesn't help me.
any help is much appreciated.
 A: This Fourier transfom exists in the distribution sense and Mathematica 12.2 finds it in analytical form by
FourierTransform[RealAbs[Sin[2 x]], x, s]

$$\sum _{K[1]=-\infty }^{\infty } \frac{\sqrt{\frac{2}{\pi }} (-1)^{K[1]+1} \left((-1)^{K[1]}+1\right) \left(1+i^{K[1]}\right)^2 \delta (s+K[1])}{K[1]^2-4} .$$
The terms of the series for $K[1]=\pm 2$ look queerly, but in fact these terms equal $0$ as
Limit[((-1)^(1 + K[1])*(1 + (-1)^K[1])*(1 + I^K[1])^2*
Sqrt[2/Pi])/(-4 + K[1]^2), K[1] -> 2]

$$ 0$$ shows.
Also the result of
InverseFourierTransform[Sum[((-1)^(1 + K[1])*(1 + (-1)^K[1])*(1 + I^K[1])^2*Sqrt[2/Pi]*
 DiracDelta[s + K[1]])/(-4 + K[1]^2), {K[1], -Infinity,Infinity}], s, x]

$$\sum _{K[1]=-\infty }^{\infty } \frac{(-1)^{K[1]+1} \left((-1)^{K[1]}+1\right) \left(1+i^{K[1]}\right)^2 e^{i x K[1]}}{\pi  \left(K[1]^2-4\right)} $$
confirms it though the one is not explicitly equal to $|\sin 2x|$. Noting again that the terms for $K[1]=\pm 2 $ equal zero, we draw the plot
Plot[Sum[((-1)^(1 + K[1])*(1 + (-1)^K[1])*(1 + I^K[1])^2*
E^(I*x*K[1]))/(Pi*(-4 + K[1]^2)), {K[1], -100, -3}] + 
Sum[((-1)^(1 + K[1])*(1 + (-1)^K[1])*(1 + I^K[1])^2*
E^(I*x*K[1]))/(Pi*(-4 + K[1]^2)), {K[1], -1, 1}] + 
Sum[((-1)^(1 + K[1])*(1 + (-1)^K[1])*(1 + I^K[1])^2*
E^(I*x*K[1]))/(Pi*(-4 + K[1]^2)), {K[1], 3, 100}], {x, 0, Pi/2}]


