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:
- Heaviside's Fourier Transform: To calculate Heaviside's Fourier transform, I am using that I know the Fourier transform of the sign function: $$\cal{F}(\text{sign})=\frac{2}{iy}.$$ So, as we can write the Heaviside function $H$ as $$H(x)=\frac{\text{sign}(x)+1}{2},$$ and using that $\cal{F}(1)=2\pi\delta$, i get:
$$\mathcal{F}(H)(y)=\frac{1}{iy}+\pi\delta$$ - Absolute Value's Fourier Transform: 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 Heaviside, like this: $|t|=tH(t)-tH(-t)$. Now, I use the relation $D(\mathcal{F}(f))=\cal{F}(-itf)$, which lets me write $$\mathcal{F}(tH(t))=\frac{1}{-i}D(\mathcal{F}(H))=\frac{1}{-i}D\Bigl(\frac{1}{iy}+\pi\delta\Bigr).$$
Taking into account that the derivative of the Dirac's delta is $D(\delta)(\varphi)=-\varphi'(0)$, I have:
$$\mathcal{F}(tH(t))(\varphi)=\Bigl(-\frac{1}{y^2}-i\pi\varphi'(0)\Bigr)$$
Now, using the fact that $\mathcal{F}(f)(ay)=|a|^{-1}\cal{F}(f)(a^{-1}y)$:
$$\mathcal{F}(-tH(-t))(\varphi)=\Bigl(-\frac{1}{(-y)^2}-i\pi\varphi'(0)\Bigr)=\Bigl(-\frac{1}{y^2}-i\pi\varphi'(0)\Bigr)$$
And combining these results I think we can say this:
$$\mathcal{F}(|t|)(\varphi)=\mathcal{F}(tH(t))(\varphi)+\mathcal{F}(-tH(-t))(\varphi)=-2\Bigl(\frac{1}{y^2}+i\pi\varphi'(0)\Bigr)$$
I'm not sure if my reasonings are correct. Thanks a lot in advance for any help!
Best Answer
I suggest to write $$|x|=-x+2xH(x)$$ knowing that $$\mathcal{F}[x] = \delta_0'$$ 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 $\mathcal{F}[H(x)](\xi)=\text{vp}(1/\xi)$ (again, up to another multiplicative constant). Now we need to divide it by $\xi$, which will give us $\text{vp}(1/\xi^2)$ and a term $C\delta_0$ with unknown $C$. However, the inverse Fourier Transform would translate as $\delta'\to x$, $\delta\to 1$, and $\text{vp}(1/\xi^2)\to$ the discontinuity of derivative in zero, so I think it's possible to show that $C=0.$
If you need help with formalising this approach, ask in comments.
Another approach would be to write $ix\,\text{sign}(x)=i|x|$, which implies that (up to some irrelevant multiplicative constant) $$\mathcal{F}[|x|](\xi) = -2 (\text{vp}(1/\xi))'.$$