The Fourier transform of a convolution
$$(f*g)(t)=\int_{-\infty}^{\infty}g(y)f(x-y)\text{d}y$$
is just $G(t)F(t)$ where $G(t),F(t)$ are the transforms of $f$ and $g$. We also have that the transform of
$$\int_{-\infty}^tf(s)\text{d}s$$
is $\frac{1}{i\omega}\tilde{f}(\omega)+\pi\tilde{f}(0)\delta(\omega)$, which can be derived using the transform of the convolution.
But what would be the transform of
$$\int_{-\infty}^{t}g(s)f(t-s)\text{d}s,$$
can it somehow be derived using the above two transformations?
Let $h(t)=\int_{-\infty}^tg(s)f(t-s)\,ds$. Then, $H(\omega)$ is
$$\begin{align} H(\omega)&=\int_{-\infty}^\infty \left(\int_{-\infty}^tg(s)f(t-s)\,ds\right)e^{i\omega t}\,dt\\\\ &=\int_{-\infty}^\infty \left(\int_{-\infty}^\infty g(s)u(t-s)f(t-s)\,ds\right)e^{i\omega t}\,dt\\\\ &=G(\omega)\mathscr{F}\{uf\}\\\\ &=G(\omega)\,\frac{1}{2\pi}\left(F(\omega)*U(\omega)\right)\\\\ &=G(\omega)\int_{-\infty}^\infty F(\omega-\omega')\left(\frac12 \delta(\omega')+\frac{i}{2\pi\omega'}\right)\,d\omega'\\\\ &=\frac12F(\omega)G(\omega)+\frac{i}{2\pi}G(\omega)\int_{-\infty}^\infty \frac{F(\omega')}{\omega-\omega'}\,d\omega' \end{align}$$