I am trying to calculate the Fourier transform $\mathcal{F}[H(t-\tau)x(t),t,\omega]$ where $H(t)$ is the Heaviside distribution and x(t) is a well behaved real function (like the position of a bead in a trap. Nothing crazy, nothing diverging, everything continuous, etc). It seems relatively straightforward but in the end, I obtain $\mathcal{F}[H(t-\tau)x(t),t,\omega]=\mathcal{F}[x(t),t,\omega]$ and can't figure out why the heaviside distribution just seems to vanish.
I'd like to understand what I do wrong and what I should do instead.
Here is my attempt :
for compactness, I will use the notation $\mathcal{F}[f(t),t,\omega] = \hat{f}(\omega)$. $$ \mathcal{F}[H(t-\tau)x(t),t,\omega]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}H(t-\tau)x(t)e^{-i \omega t}dt \\ \mathcal{F}[H(t-\tau)x(t),t,\omega]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{-i (\omega-\omega_1)\tau}\hat{H}(\omega-\omega_1)\hat{x}(\omega_1)d\omega_1 $$ where I used both the product rule and the translation rule.
I then use the fact that $\hat{H}(\omega)=\sqrt{\frac{\pi}{2}}\left(\frac{1}{i \pi \omega}+\delta(\omega)\right)$ and separate the dirac part of the integral, which is pretty trivial. $$ \mathcal{F}[H(t-\tau)x(t),t,\omega]=\frac{1}{2}\hat{x}(\omega)+\frac{e^{-i \omega \tau}}{2 i \pi}\int_{-\infty}^{+\infty}\frac{\hat{x}(\omega_1)e^{i \omega_1 \tau}}{\omega-\omega_1}d\omega_1 $$
Now I have to compute this integral that has a pole on $\omega$. To do this, I have to use the imaginary half plane because of the $e^{i \omega_1 \tau}$ term. I can then write the integral on a closed loop containing no poles : $$ \int_{\omega+\epsilon}^R \frac{\hat{x}(\omega_1)e^{i \omega_1 \tau}}{\omega-\omega_1} d\omega_1 + \int_0^\pi \frac{\hat{x}(R e^{i \theta})e^{i R e^{i \theta} \tau}}{\omega-R e^{i \theta}} i R e^{i \theta} d\theta + \int_{-R}^{\omega-\epsilon} \frac{\hat{x}(\omega_1)e^{i \omega_1 \tau}}{\omega-\omega_1} d\omega_1 + \int_\pi^0 \frac{\hat{x}(\omega + \epsilon e^{i \theta})e^{i (\omega + \epsilon e^{i \theta}) \tau}}{\omega-(\omega + \epsilon e^{i \theta})} i \epsilon e^{i \theta} d\theta = 0 $$ when $R\to\infty$ and $\epsilon\to 0$ : $$ \int_0^\pi \frac{\hat{x}(R e^{i \theta})e^{i R e^{i \theta} \tau}}{\omega-R e^{i \theta}} i R e^{i \theta} d\theta \to 0 \\ \int_{-R}^{\omega-\epsilon} \frac{\hat{x}(\omega_1)e^{i \omega_1 \tau}}{\omega-\omega_1} d\omega_1 + \int_{\omega+\epsilon}^R \frac{\hat{x}(\omega_1)e^{i \omega_1 \tau}}{\omega-\omega_1} d\omega_1 \to \int_{-\infty}^{+\infty} \frac{\hat{x}(\omega_1)e^{i \omega_1 \tau}}{\omega-\omega_1} d\omega_1 $$ The remaining term is $$ \int_\pi^0 \frac{\hat{x}(\omega + \epsilon e^{i \theta})e^{i (\omega + \epsilon e^{i \theta}) \tau}}{\omega-(\omega + \epsilon e^{i \theta})} i \epsilon e^{i \theta} d\theta = -\int_0^\pi \hat{x}(\omega + \epsilon e^{i \theta})e^{i (\omega + \epsilon e^{i \theta}) \tau} i d\theta \\ \int_\pi^0 \frac{\hat{x}(\omega + \epsilon e^{i \theta})e^{i (\omega + \epsilon e^{i \theta}) \tau}}{\omega-(\omega + \epsilon e^{i \theta})} i \epsilon e^{i \theta} d\theta \to -i \int_0^\pi \hat{x}(\omega)e^{i \omega \tau} d\theta = -i\pi \hat{x}(\omega)e^{i \omega \tau} $$
In the end, we can rewrite this integral as $\int_{-\infty}^{+\infty} \frac{\hat{x}(\omega_1)e^{i \omega_1 \tau}}{\omega-\omega_1} d\omega_1 = i\pi \hat{x}(\omega)e^{i \omega \tau}$. Therefore my equation becomes : $$ \mathcal{F}[H(t-\tau)x(t),t,\omega]=\frac{1}{2}\hat{x}(\omega)+\frac{e^{-i \omega \tau}}{2 i \pi} i\pi \hat{x}(\omega)e^{i \omega \tau} = \hat{x}(\omega) $$
If I then take the inverse Fourier transform of this , I obviously do not come back to what I started with ... So obviously this has to be wrong, but I cannot understand where I go wrong.
What is false in here, and what should I do to get the right result ? (and what is the right result, by the way ?)
The Fourier transform of $H(t)x(t)$ is $A \hat{x}(\omega) + \hat{x} \ast pv(\frac{B}{\omega})$ and the second term doesn't simplify in general. If $\hat{x}$ is analytic, to evaluate $\int_{-\infty}^\infty \hat{x}(w) \frac{1}{\omega-w}dw$ as a contour integral you need to take the mean value of two integrals $$\frac12 (\int_{C^+} \hat{x}(w) \frac{1}{\omega-w}dw+\int_{C^-} \hat{x}(w) \frac{1}{\omega-w}dw)$$ where $C^+,C^-$ are the real axis with a small indentation above/below $w = \omega$.
If $\hat{x}(z)$ decreases fast enough as $\Im(z) \to \infty$ the residue theorem applies and you get $$\int_{-\infty}^\infty \hat{x}(w) \frac{1}{\omega-w}dw =\frac12 \int_{C^-} \hat{x}(w) \frac{1}{\omega-w}dw= i \pi Res(\hat{x}(w) \frac{1}{\omega-w}, \omega) = -i\pi \hat{x}(\omega)$$ indicating that $x$ was indeed supported on $t > 0$.
If instead $x$ was supported on $t< 0$ it is $$\int_{-\infty}^\infty \hat{x}(w) \frac{1}{\omega-w}dw =\frac12 \int_{C^+} \hat{x}(w) \frac{1}{\omega-w}dw= -i \pi Res(\hat{x}(w) \frac{1}{\omega-w}, \omega) = i\pi \hat{x}(\omega)$$ as expected from the fact that $pv(\frac{1/i\pi}{\omega})$ is the Fourier transform of $sign(t)$.