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I'm trying to obtain these two Fourier transforms. First of all, I'm using the following definition of Fourier transform:

$$\cal{F}(f)(y)=\int_{-\infty}^{\infty}f(x)e^{-ixy}\;dx$$


What I have so far is this:

To calculate Heavyside's Fourier transform, I am using that I know the Fourier transform of the sign function: $\displaystyle\cal{F}(sign)=\frac{2}{iy}$. So , as we can write The Heavyside function $H$ as $\displaystyle H(x)=\frac{sign(x)+1}{2}$, and using that $\cal{F}(1)=2\pi\delta$, i get:

$$\cal{F}(H)(y)=\frac{1}{iy}+\pi\delta$$

To get the Fourier transform of the absolute value function, I'm trying to use that fact that we can write it in terms of the Heavyside, like this: $|t|=tH(t)-tH(-t)$.

Now, I use the relation $D(\cal{F}(f))=\cal{F}(-itf)$, which lets me write $\cal{F}(tH(t))=\frac{1}{-i}D(\cal{F}(H))=\frac{1}{-i}D(\frac{1}{iy}+\pi\delta)$.

Taking into account that the derivative of the Dirac's delta is $D(\delta)(\varphi)=-\varphi'(0)$, I have:

$$\cal{F}(tH(t))(\varphi)=(-\frac{1}{y^2}-i\pi\varphi'(0))$$

Now, using the fact that $\cal{F}(f)(ay)=|a|^{-1}\cal{F}(f)(a^{-1}y)$:

$$\cal{F}(-tH(-t))(\varphi)=(-\frac{1}{(-y)^2}-i\pi\varphi'(0))=(-\frac{1}{y^2}-i\pi\varphi'(0))$$

And combining these results I think we can say this:

$$\cal{F}(|t|)(\varphi)=\cal{F}(tH(t))(\varphi)+\cal{F}(-tH(-t))(\varphi)=-2(\frac{1}{y^2}+i\pi\varphi'(0))$$


I'm not sure if my reasonings are correct. Thanks a lot in advance for any help!

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I realized that the last part, where I try to compute $\cal{F}(-tH(-t))(\varphi)$ is wrong. In essence, I would need some formula to compute the Fourier transform of the 'check' of a function (changing the sign of the variable). Any ideas? –  Mark_Hoffman Jan 23 at 19:45
    
Strangely enough, Mathematica insists that the coefficient at $\delta'$ is zero. –  TZakrevskiy Jan 26 at 14:30

1 Answer 1

I suggest to write $$|x|=-x+2xH(x).$$

$$\mathcal F[x] = \delta_0'$$ up to a multiplicative constant that depends on the notation convention for Fourier Transform.

Then again, to find $\mathcal F[xH(x)]$ one can use that fact that $(xH(x))'= H(X)$ and that $\mathcal F[H(x)](\xi)=vp(1/\xi) $ (again, up to another multiplicative constant). Now we need to divide it by $\xi$, which will give us $vp(1/\xi^2)$ and a term $C\delta_0$ with unknown $C$. However, the inverse Fourier transform would translate as $\delta'\to$ x, $\delta\to$ 1, and $vp(1/\xi^2)\to$ the discontinuity of derivative in zero, so I think it's possibble to show that $C=0.$

If you need help with formalising this approach, ask in comments.

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