How to compute the Fourier transform of this function?

Question

In our course, the Fourier transform of $f\in L^1(\mathbb{R})$ is defined as, $$\mathcal{F}(f) = \hat{f}(\zeta) = \int e^{-ix\zeta}f(x)dx.$$ Therefore, the inverse fourier transform is defined as, $$\bar{\mathcal{F}}(f) = f(x) =\frac{1}{2\pi} \int e^{ix\zeta} \hat{f}(\zeta)d\zeta.$$ Furthermore, we know that, $$\mathcal{F}\left(\frac{2\alpha}{\alpha^2+x^2}\right) = 2\pi e^{-\alpha|\zeta|}.$$ My goal is to compute $$\mathcal{F}\left(\frac{1}{x+i}\right).$$ Here is what I have tried so far. For $\alpha =1$ we get that, $$\mathcal{F}\left(\frac{1}{x+i}\cdot \frac{1}{x-i}\right)=\frac{1}{2\pi}\mathcal{F}\left(\frac{1}{x+i}\right)*\mathcal{F}\left(\frac{1}{x-i}\right) =\pi e^{-|\zeta|}.$$ Now if $$\hat{f}(\zeta) = \mathcal{F}\left(\frac{1}{x+i}\right)$$ then, $$\mathcal{F}\left(\frac{1}{x-i}\right) = \hat{f}(-\zeta).$$ Thus we want to solve $$\hat{f}(x)*\hat{f}(-x) = 2\pi^2 e^{-|\zeta|}.$$ I am not sure how to proceed after this step. Any hints will be much appreciated.