Evaluate $\langle \mathbf{p} | 1/\hat{r} | \mathbf{p}' \rangle$ In Sakurai's Problem 1.27 b), we use $\langle \mathbf{r} | \mathbf{p}\rangle = e^{i\mathbf{p}\cdot\mathbf{r}/\hbar}$ to show that 
$$
\langle \mathbf{p} | F(\hat{r}) | \mathbf{p}' \rangle = \frac{1}{2 \pi^2 \hbar^2 q} \int_{0}^{\infty}r'\sin(q r'/\hbar) F(r') \,\text{d}r',
$$
where $q = |\mathbf{p} - \mathbf{p}'|$. So if I just set $F(r) = 1/r$, then 
$$
\langle \mathbf{p} | F(\hat{r}) | \mathbf{p}' \rangle = \frac{1}{2 \pi^2 \hbar^2 q} \int_{0}^{\infty}\sin(q r'/\hbar) \,\text{d}r',
$$
but that clearly doesn't converge as it is. So, is that just it?
 A: Consider the general case that we want to calculate $$
\langle p |F(r) |p'\rangle.$$
By inserting the resolution of the identity $\int d^3r\, |r\rangle\langle r|$ we find that we need to compute $$\tilde{F}(q = p-p') = \int d^3 r \, e^{i(p-p')r} F(r). \tag{1}$$
This integral will converge if $\int dr\, |F(r)|$ is finite. Such a function is said to be $L^1$. It will also be the case that $\int dq\, | \tilde F (q)|$ is finite, that is, $\tilde F(q)$ is also $L^1$. The map $F \mapsto \tilde F$ is the Fourier transform, let's call it $\mathcal F$.
Now the problem is that $1/r$ in 3-dimensional space is not $L^1$, but if we can find a way to extend the Fourier transform to be defined for more functions, maybe we can still make sense of $\langle p |\frac{1}{r}|p' \rangle$. The Fourier transform can be extended to functions that satisfy only $$\int d^3 r \, |F(r) g(r)| < \infty \tag{2}$$
for all functions $g(r)$ that decay rapidly enough (more precisely, $g$ needs to decay faster than any power of $r$). However by extending this far, $\tilde F(q)$ isn't necessarily $L^1$, or even a function anymore. It could be a distribution, like a Dirac delta. Maybe you have seen the formula $$\int d^3r\, e^{i (\vec p - \vec p') \cdot \vec r} = \delta(\vec p- \vec p').$$
What is really meant is that $$(\mathcal F[1])(p) = \delta(p).$$
Connecting back to the original question, the function $F(r) = 1/r$ satisfies the condition (2), but we cannot use the integral formula (1) to find $\tilde F(q)$. However, the functions $F_\epsilon(r) = e^{-\epsilon r}/r$, with $\epsilon > 0$, converge to $F$ when $\epsilon \to 0$. Each $F_\epsilon(r)$ is $L^1$, so we can (rather) easily find $\tilde F_\epsilon(q)$. The extension of the Fourier transform is so constructed that if $$F_\epsilon \overset{\epsilon \to 0}{\longrightarrow} F \text{ then } \tilde{F}_\epsilon \overset{\epsilon \to 0}{\longrightarrow} \tilde F.$$
This justifies using the convergence factor $e^{-\epsilon r}$.
(You are right that you can make almost anything converge by damping with $e^{-\epsilon r}$, but that just reflects that since the condition (2) isn't very strong, the Fourier transform is defined for a whole bunch of functions.)
