Fourier transform of time division
A unit step in the frequency domain $u(f)$ has the following inverse Fourier transform:
$$\mathcal{F}^{-1}\{u(f)\}=\frac12\delta(t)-\frac{1}{2\pi jt}\tag{1}$$
This follows from the well-known Fourier transform of the time-domain unit step $u(t)$, and from the basic Fourier transform property
$$\mathcal{F}\{x(t)\}=X(f)\Longrightarrow\mathcal{F}^{-1}\{x(f)\}=X(-t)\tag{2}$$
Given $x(t)$ and its Fourier transform $X(f)$, we have
$$(X\star u)(f)=\int_{-\infty}^fX(\nu)d\nu\tag{3}$$
where $\star$ denotes convolution.
Using $(1)$, the inverse Fourier transform of $(3)$ is
$$x(t)\cdot \mathcal{F}^{-1}\{u(f)\}=\frac{x(0)}{2}\delta(t)-\frac{x(t)}{2\pi jt}\tag{4}$$
Consequently,
$$\mathcal{F}\left\{\frac{x(t)}{t}\right\}=2\pi j\left[\frac{x(0)}{2}-\int_{-\infty}^fX(\nu)d\nu\right]\tag{5}$$
As an exercise, you can try to verify $(5)$ with $x(t)=\sin(2\pi f_0t)/\pi$ (resulting in an ideal low-pass with cut-off frequency $f_0$), and with $x(t)=1/\pi$ (resulting in an ideal Hilbert transformer).