Proving $\mathcal{F}^{-1}\left\{\frac{i}{2t}\hat{f}(t)\right\}=-\int_{-\infty}^tf(s)\,ds$ in the sense of distributions Let $f\in C^\infty_0(\mathbb{R})$ with $\operatorname{supp}f\subset(0,\infty)$. I would like to prove that

$$\mathcal{F}^{-1}\left\{\frac{i}{2t}\hat{f}(t)\right\}=-\int_{-\infty}^tf(s)\,ds,\qquad t\in\mathbb{R}\qquad(\star)$$

where $c>0$ is a constant, using rigorous distribution theory
"Unrigorous" Proof:
The approach I considered is to consider the fact that the Fourier transform of the Heaviside function is given by
$$\hat{H}(\omega)=\frac{1}{2}\delta(\omega)-\frac{i}{2\pi t}.$$
Hence, we may write
$$
\begin{aligned}
\mathcal{F}^{-1}\left\{\frac{i}{2t}\hat{f}(t)\right\}&=\pi\mathcal{F}^{-1}\left\{\frac{i}{2\pi t}\hat{f}(t)\right\}
\\
&=-\pi\mathcal{F}^{-1}\{\hat{H}(\omega)\hat{f}(t)\}\qquad(\omega\ne 0)
\end{aligned}$$
where we have used that $0\notin\operatorname{supp}f$ and $\operatorname{supp}\delta=\{0\}$.
Then $(\star)$ follows via an application of the convolution theorem.
Now, my question is:

Can we rigorously prove $(\star)$ in the sense of distributions

 A: Using $\hat{H}(\xi ) = pv.(\frac{1}{2 i\pi \xi})+\frac{1}{2}  \delta(\xi)$ and the convolution theorem
$$\forall \phi,f\in S(\mathbb{R}), \qquad  \langle pv.(\frac{\hat{f}(\xi)}{2 i\pi \xi})+\frac{\hat{f}(0)}{2}  \delta(\xi), \hat{\phi}\rangle = \langle \hat{H}, \hat{f}\hat{\phi}\rangle = \langle H, f \ast \phi\rangle =\langle H \ast f, \phi\rangle $$
Qed. if $F(x) =H \ast f(x)= \int_{-\infty}^x f(y)dy$ then
$$ \hat{F}(\xi) = pv.(\frac{\hat{f}(\xi)}{2 i\pi \xi})+\frac{\hat{f}(0)}{2}  \delta(\xi)$$
A: As definition of the Fourier transform we take
$$\hat\phi(\xi) = \int \phi(x) e^{-i\xi x} \, dx$$
Then we have $\hat\delta = 1$ since $\langle \hat\delta(t), \phi(t) \rangle = \langle \delta(t), \hat\phi(t) \rangle = \hat\phi(0) = \int \phi(x) 1 \, dx = \langle 1(t), \phi(t) \rangle.$
Now, $H = \frac12 (1 + \theta),$ where $\theta(x) = -1$ when $x<0$, and $\theta(x) = +1$ when $x>0$.
For the constant part we have $\hat 1(t) = \hat{\hat\delta}(t) = 2\pi \, \delta(-t) = 2\pi \, \delta(t).$
For $\theta$ we have $\theta' = 2\delta$ so $it \hat\theta(t) = \widehat{\theta'}(t) = \hat\delta(t) = 1(t).$ Thus $\hat\theta(t) = -i \, \operatorname{pv}\cfrac{1}{t}$.
Summarizing we get
$$\hat H(t) = \frac12 (\hat 1 + \hat\theta) = \frac12 \left(2\pi\,\delta(t) - i\,\operatorname{pv}\cfrac{1}{t}\right)$$

If we instead define the Fourier transform as
$$\hat\phi(\xi) = \frac{1}{2\pi} \int \phi(x) e^{-i\xi x} \, dx$$
we get
$$\hat H(t) = \frac12 \left(\delta(t) - i \frac{1}{2\pi} \,\operatorname{pv}\cfrac{1}{t}\right)$$
There's a small difference from the formula that Jason Born had. Have I made a mistake or is it an error in Jason's formula?
