Fourier transform of $\frac{d}{dt}\ln\frac{1}{it}$

Question

I'd like to proove the identity

$$\mathcal{F}\left(\frac{d}{dt}\ln\frac{1}{it}\right)=2i\pi H$$

with $H=\mathbb{I}_{\mathbb{R}^+}$ ie the Heaviside step function, $\mathcal{F}$ denote the Fourier transform on distributions and $\ln$ the principal logarithme value. If it's correct I make a mistake somewhere but don't find where:

I get $\mathcal{F}\left(\operatorname{vp}\frac{1}{x}\right)=-2i\pi H+i\pi$: correct?

One has $\ln\left(\frac{1}{it}\right)=\ln\left|\frac{1}{it} \right|+i\arg\left(\frac{1}{it} \right)=-\ln|t|-i\frac{\pi}{2}\operatorname{sign}(t)$.

So we deduce, $\frac{d}{dt}\ln\frac{1}{it}=-\operatorname{vp}\frac{1}{t}-i\pi\delta$.

So we conclude

$$ \mathcal{F}\left(\frac{d}{dt}\ln\frac{1}{it}\right)=-\mathcal{F}\left( \operatorname{vp}\frac{1}{t} \right)-\mathcal{F}(i\pi\delta)=-\left( -2i\pi H+i\pi \right)-i\pi\times1=2i\pi H-2i\pi$$

which is not what I guess: where is my problem? Very probably a stupid sign problem somewhere...

Answer 1

You have $$ \mathcal{F}(\text{ vp}\frac{1}{t}) = -i\pi\text{ sgn } = 2i\pi H(-\cdot) -i\pi $$ and hence $$ \mathcal{F}(\frac{d}{dt}\text{ ln}\frac{1}{it}) = -2 i \pi H(-\cdot), $$ so for a test function $\varphi$ $$ \mathcal{F}(\frac{d}{dt}\text{ ln}\frac{1}{it})(\varphi) = -2i\pi \int_{-\infty}^0 \varphi(t) dt $$ which is exactly the same as you calculated, but not expected.