Integrating An Absolute Value Using Jordan's Lemma? I am working with fermion field expansions in QFT. The fields are given by 
$$\psi(x)=\int d\mathbf{k}\frac{1}{\sqrt{2E_k}}[b_{k}u(k)e^{-ikx}+d_k^{\dagger}v(k)e^{ikx}]$$
I am evaluating 
$$\int d^3x d^3x' \bar{\psi(x)}\psi(x)\frac{1}{|\mathbf{x-x'}|}\bar{\psi(x')}\psi(x')$$
 which when expanded gives me 
$$\begin{multline}
\int d^3x d^3x' d\mathbf{k}d\mathbf{k'}d\mathbf{p}d\mathbf{p'}\frac{1}{\sqrt{2E_kE_k'E_pE_p'}} \\ [b_{k}u(k)e^{-ikx}+d_k^{\dagger}v(k)e^{ikx}][b_{k'}u(k')e^{-ik'x}+d_{k'}^{\dagger}v(k')e^{ik'x}]\times\frac{1}{|\mathbf{x}-\mathbf{x'}|}[b_{p}u(p)e^{-ipx'}+d_p^{\dagger}v(p)e^{ipx'}][b_{p'}u(p')e^{-ip'x'}+d_{p'}^{\dagger}v(p')e^{ip'x'}]\end{multline}$$
This gives me integrals like the following to evaluate:
$$\int d^3x\frac{e^{i\mathbf{(k-k')x}}}{|\mathbf{x}-\mathbf{x'}|}$$
I was wondering if it is possible to evaluate this integral using Jordan Lemma. Alternatively, I have evaluated it partially by computing the angular part first, that is, by writing
$$\int d^3x\frac{e^{i\mathbf{(k-k')x}}}{|\mathbf{x}-\mathbf{x'}|} = \int^{\infty}_{0} \int^{2\pi}_{0} \int^{\pi}_{-\pi}dx x^2 d\phi d\theta \frac{e^{i(k-k')x\cos{\theta}}}{|x^2+x'^{2}-xx'\cos{\alpha}|}$$
This gives me
$$\frac{\pi}{(k-k')}\int^{\infty}_{0}dx\frac{x\sin((k-k')x)}{(x^2+x'^2-2xx'\cos\alpha)^{1/2}}$$
Can I use complex variables to integrate this last integral? Because I don't know how to integrate it with usual methods. Or is there another way?
Thank you in advance!
 A: Use a change of variables to shift the origin to $\mathbf{x'}$:
$$\int d^3 \mathbf{x} \frac{e^{i(\mathbf{k-k'})\mathbf{x}}}{|\mathbf{x-x'}|} = e^{i(\mathbf{k-k'})\mathbf{x'}}\int d^3 \mathbf{x} \frac{e^{i(\mathbf{k-k'})\mathbf{x}}}{|\mathbf{x}|} = e^{i(\mathbf{k-k'})\mathbf{x'}} \mathcal{F}\left\{\frac{1}{|\mathbf{x}|}\right\}(\mathbf{k-k'})$$
using the $\mathcal{F}\{f(\mathbf{x})\}(\mathbf{k}) = \int d^3 \mathbf{x} f(x)e^{i\mathbf{kx}}$ convention. There are several known tricks to evaluate this Fourier transform, but the most elegant involves the property of homogeneity, as I discuss here. 
Evaluating the constant gives us the area of the unit sphere, $4\pi$, giving that
$$\mathcal{F}\left\{\frac{1}{|\mathbf{x}|}\right\}(\mathbf{k}) = \frac{4\pi}{|\mathbf{k}|^2}$$
(letting $\mathbf{x} \mapsto \mathbf{-x}$ does not change this result). This means the integral evaluates to
$$\int d^3 \mathbf{x} \frac{e^{i(\mathbf{k-k'})\mathbf{x}}}{|\mathbf{x-x'}|} = \frac{4\pi e^{i(\mathbf{k-k'})\mathbf{x'}} }{|\mathbf{k-k'}|^2}$$
