$\renewcommand{div}{\mathop{\mathrm{div}}}$ Let $a(x) \colon \mathbb R^n \to \mathbb R^n$ be a smooth vector field with compact support. Let $\hat a(\xi)$ be it's Fourier transform: $$ \hat a(\xi) = \int\limits_{\mathbb R^n} \exp(-i\xi x) a(x) \, dx. $$ I need to find the inverse Fourier transform of the field $\left(\hat a(\xi) \cdot \frac{\xi}{|\xi|} \right) \frac{\xi}{|\xi|}$ that is a projection of $\hat a(\xi)$ on the direction $\frac{\xi}{|\xi|}$, $\xi \neq 0$.
We have $$ \int\limits_{\mathbb R^n} \exp(-i\xi x) \div a(x) \, dx = \int\limits_{\mathbb R^n} \div\left[ \exp(-i\xi x) a(x) \right] \, dx + i\xi \hat a(\xi) \\ = \lim\limits_{R \to + \infty} \int\limits_{|x|=R} \exp(-i\xi x) a(x)\cdot \frac{x}{|x|} \, dx + i\xi \hat a(\xi) = i\xi \hat a(\xi). $$ Then $\widehat{\div a}(\xi) = i\xi \hat a(\xi)$ and $$ \left(\hat a(\xi) \cdot \frac{\xi}{|\xi|} \right) \frac{\xi}{|\xi|} = -i\widehat{\div a}{\xi} \frac{\xi}{|\xi|^2}. $$ Similarly it can be shown that $i \widehat{\div a}(\xi) \xi = -(\nabla \div a)^\wedge (\xi)$ so that $$ \left(\hat a(\xi) \cdot \frac{\xi}{|\xi|} \right) \frac{\xi}{|\xi|} = \frac{(\nabla \div a)^\wedge (\xi)}{|\xi|^2}. $$ So the question is what to do with $|\xi|^2$ in the denominator to find the explicit expression for $$ c(x) = (2\pi)^{-n} \int\limits_{\mathbb R^n} \exp(i\xi x) \left(\hat a(\xi) \cdot \frac{\xi}{|\xi|} \right) \frac{\xi}{|\xi|} \, d\xi = (2\pi)^{-n} \int\limits_{\mathbb R^n} \exp(i\xi x) \frac{(\nabla \div a)^\wedge (\xi)}{|\xi|^2} \, d\xi $$ in terms of $a(x)$ and it's derivatives?
Since $\Delta \exp(i\xi x) = -|\xi|^2 \exp(i\xi x)$ we can see that $\Delta c(x) = - \nabla \div a(x)$. I don't know if this can help.