How to integrate $I(k)=\frac{2\pi}{i k^2}\int_0^{\infty}\left(e^{-ir }-e^{ir }\right)dr$ I heard that you can integrate 
$$\begin{align}I(k)=\frac{2\pi}{i k^2}\int_0^{\infty}\left(e^{-ir }-e^{ir }\right)dr
\end{align}$$ in the sense of tempered distribution. Unfortunately, I am only familiar with "standard integration", but I need this integral for a physics exercise which is why I wanted to ask here if anybody could show me, how such an integral is calculated?
 A: If a physicist encounters an integral in which a physical variable runs to infinity, and if for that very reason the integral is not strictly convergent in a precise mathematical sense, he will usually solve the problem by inserting a so-called "convergence factor" in the integrand. This is a function that is nearly equal to 1 in the normal range of (physical) values but drops off sufficiently for very large values to make the integration convergent. This can almost always be justified, on the grounds that in reality physical variables tend to have some upper limit, a cut-off value determined by the physical system. 
The most obvious choice in your case is $f(r) = \exp(-\epsilon r)$. That is because performing the integration is then straightforward. At the end of the computation, one will usually take the limit of $\epsilon\to 0$. Note that other choices for the convergence factor are probably equally valid, e.g. a Gaussian, $f(r) = \exp(-(\epsilon r)^2)$.  
A: I should have finished my answer on your original question even though there was an accepted answer, sorry for that. Seeing that things are not yet clear for you, I chose to post it here instead.
The Fourier transformation $\mathcal{F}$ of a tempered distribution $u \in \mathcal{S}'$ is defined as the "adjoint" of the Fourier transformation on the Schwartz space $\mathcal{S}$, that is $\mathcal{F}u:=u \circ \mathcal{F}$ (or $\forall \phi \in \mathcal{S}: (\mathcal{F}u)(\phi)=u(\mathcal{F}(\phi))$ if you wish). You are given the tempered distribution $u_{\frac{1}{\vert x \vert}} \in \mathcal{S}'(\mathbb{R}^{3})$ defined through $u_{\frac{1}{\vert x \vert}}(\phi):=\int\limits_{\mathbb{R}^{3}}\frac{1}{\vert x \vert} \phi(x) \, \text{d}x$. By dominated convergence, $(\mathcal{F}u_{\frac{1}{\vert x \vert}})(\phi)=\lim\limits_{\epsilon \searrow 0} \int\limits_{\mathbb{R^{3}}}\frac{1}{\vert x \vert} e^{-\epsilon \vert x \vert} (\mathcal{F}\phi)(x) \, \text{d}x$. But for $\epsilon > 0$, the function $x \mapsto \frac{1}{\vert x \vert} e^{-\epsilon \vert x \vert}$ lies in $L^{1}(\mathbb{R}^{3}) \cap L^{2}(\mathbb{R}^{3})$, so we may use the direct consequence of Fubini's theorem that $\int f (\mathcal{F}\phi) = \int (\mathcal{F}f) \phi$ for certain functions and compute the Fourier transformation of $\frac{1}{\vert x \vert} e^{-\epsilon \vert x \vert}$ in the usual sense.
For $k \neq 0$ ($k=0$ is easily done), pick $R \in \text{O}(3)$ with $Rk=\vert k \vert e_{3}$. Substituting $x=R^{t}y$ and then changing to polar coordinates gives us for the Fourier transformation
$$\begin{aligned}\int\limits_{\mathbb{R^{3}}}\frac{1}{\vert x \vert} e^{-\epsilon \vert x \vert+i \langle k, x \rangle } \, \text{d}x &= \int\limits_{\mathbb{R^{3}}}\frac{1}{\vert y \vert} e^{-\epsilon \vert y \vert+i \vert k \vert y_{3} } \, \text{d}y = \int\limits_{0}^{\infty}\int\limits_{0}^{2 \pi}\int\limits_{0}^{\pi} e^{-\epsilon r+i \vert k \vert r \cos(\theta) }  r \sin(\theta)\, \text{d}\theta \, \text{d}\phi \, \text{d}r \\ &= 2 \pi \int\limits_{0}^{\infty}\int\limits_{0}^{\pi} e^{-\epsilon r+i \vert k \vert r \cos(\theta) }  r \sin(\theta) \, \text{d}\theta\, \text{d}r \stackrel{z=cos(\theta)}{=} 2 \pi \int\limits_{0}^{\infty}\int\limits_{-1}^{1} e^{-\epsilon r+i \vert k \vert r z }  r  \, \text{d}z \, \text{d}r \\ &= \frac{2 \pi}{i \vert k \vert} \int\limits_{0}^{\infty}(e^{-\epsilon r+i \vert k \vert r }-e^{-\epsilon r-i \vert k \vert r })   \, \text{d}r = \frac{2 \pi}{i \vert k \vert} \left( \frac{1}{\epsilon - i \vert k \vert} -   \frac{1}{\epsilon + i \vert k \vert}\right) \\ &= \frac{4 \pi}{\epsilon^{2}+ \vert k \vert^{2}},\end{aligned}$$
which also holds for $k=0$.
Inserting this into the definition of the Fourier transformation of tempered distributions and using dominated convergence again, we conclude that for all the Schwartz functions $\phi$ we have
$$(\mathcal{F}u_{\frac{1}{\vert x \vert}})(\phi)=\int\limits_{\mathbb{R^{3}}}\frac{1}{\vert x \vert} (\mathcal{F}\phi)(x) \, \text{d}x =4 \pi \int\limits_{\mathbb{R^{3}}}\frac{1}{\vert x \vert^{2}} \phi(x) \, \text{d}x = u_{\frac{4 \pi}{ \vert x \vert^{2}}}(\phi) $$
or, loosely speaking, $$\mathcal{F}\left(\frac{1}{\vert x \vert}\right)(k)=\frac{4 \pi}{ \vert k \vert^{2}}.$$
