problem with Fourier transform
Since

$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{i\omega x}d\omega$$

you are looking for the inverse transform of

$$G(\omega)=\frac{F(\omega-a)}{i\omega}$$

If you look at a table of Fourier transform correspondences you will find

$$\frac{F(\omega)}{i\omega}\Longleftrightarrow \int_{-\infty}^x f(y)dy$$

and

$$F(\omega-a)\Longleftrightarrow e^{iax}f(x)$$

Consequently, the function whose Fourier transform is $G(\omega)$ is the integral of a modulated version of $f(x)$:

$$G(\omega)\Longleftrightarrow \int_{-\infty}^x e^{iay}f(y)dy$$

EDIT:

The correspondence

$$\frac{F(\omega)}{i\omega}\Longleftrightarrow \int_{-\infty}^x f(y)dy\tag{1}$$

assumes that

$$F(0)=\int_{-\infty}^{\infty} f(x)dx=0$$

This assumption is often made implicitly because it is a standard requirement (obviously, a sufficient one) for the Fourier transform to exist. If

$$g(x)=\int_{-\infty}^x f(y)dy$$

it simply means that $g(x)$ is a well-behaved function satisfying $\lim_{x\rightarrow\infty}g(x)=0$. However, it is also possible to compute the Fourier transform of functions that do not satisfy this condition. In this case, the correspondence (1) has to be modified:

$$\frac{F(\omega)}{i\omega}+\pi F(0)\delta(\omega)\Longleftrightarrow \int_{-\infty}^x f(y)dy\tag{2}$$

which obviously simplifies to (1) if $F(0)=0$. If this condition is not satisfied, the result for the inverse transform of $G(\omega)$ given above must be modified by an additive constant:

$$G(\omega)\Longleftrightarrow \int_{-\infty}^x e^{iay}f(y)dy-\frac{1}{2}\int_{-\infty}^{\infty}e^{iax}f(x)dx$$