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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.

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hermi
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  • It must be a distribution. – copper.hat Apr 16 '21 at 03:23
  • It seems unlikely to have a closed form solution – Nick Alger Apr 16 '21 at 03:25
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    What is it about the absolute value that bothers you? Can't you always take $\int_a^b|f(x)|,dx$ and split it into pieces where $f(x)\ge0$ and pieces where $f(x)\le0$? – Gerry Myerson Apr 16 '21 at 03:29
  • @GerryMyerson But this doesn't contradict the definition of Fourier transform?from x = -inf to x = +inf – hermi Apr 16 '21 at 09:14
  • @NickAlger You mean the problem is not solved? – hermi Apr 16 '21 at 09:15
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    I don't know how you contradict a definition. But anyway it splits the integral into an infinity of pieces, so you get an infinite series. You might have to take it as a limit as $a\to-\infty$ and $b\to\infty$. It's worth a try, if you don't have a better idea. – Gerry Myerson Apr 16 '21 at 11:58

1 Answers1

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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}]

enter image description here

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