Laplacian of $1/r$ in a tensor As we know the  $$\nabla^2(1/r) =- 4 \pi \delta^3(r).$$ However, I recently was readling an hydrodynamic book (An introduction to dynamics of colloids By J.K.G Dhont). The Oseen tensor is defined as:
$$T(r) = \frac{1}{8\pi\eta_0 r} ( \textbf{I}+ \textbf{r r}/r^2), \quad r\neq 0. \tag{equation 5.28, page 243}$$
And then it is mentioned that: 
$$\nabla^2T(r) = \frac{1}{4\pi\eta_0 r^3} (\textbf{I}- 3\textbf{r r}/r^2), \quad r\neq 0. \tag{equation 5.118, page 284}$$
Could you tell me how we are not getting this delta dirac here? I think it is becasue of $\textbf{I}$, which is a unitary matrix. I really appreciate if you show me the calculation for the second part as well. How should we proceed the $\nabla^2$ operator on $\textbf{r r}/r^2$ which is $\hat{r} \hat{r}$?
PS: For clarification, $\textbf{r}$ is vector, so $\textbf{r}/r = \hat{r}$
 A: Because you are ignoring the origin, the Dirac delta does not appear in the expression. For evaluating derivatives of the Oseen tensor, it is helpful if you use "mixed" coordinates and define the Oseen tensor in the following way using index notation:
$$\mathcal{G}_{ij} = \frac{\delta_{ij}}{r} + \frac{x_i x_j}{r^3}$$
where I have dropped the constant $8 \pi \eta$ coefficient for convenience.
If you note that $\frac{\partial r}{\partial x_k} = \frac{x_k}{r}$, then you can differentiate sequentially applying index rules to get:
$$\mathcal{G}_{ij,k} = \frac{-\delta_{ij}x_k}{r^3} + \frac{\delta_{ik}x_j + \delta_{jk}x_i}{r^3} - \frac{3 x_i x_j x_k}{r^5}$$
$$\nabla^2\mathcal{G}_{ij} = \mathcal{G}_{ij,kk}=\frac{2\delta_{ij}}{r^3} - \frac{6x_i x_j}{r^5}$$
A: OP's differentiation formulas (including OP's well-known first identity) can of course be understood pointwise on $\mathbb{R}^3\backslash\{0\}$ where the functions are smooth. The interesting non-trivial question is whether they can be promoted to distributions on the full space $\mathbb{R}^3$? Well, let's see.
If we regularize OP's tensor (5.28) as a smooth function$^1$
$$ T_{ij}~=~\frac{\delta_{ij}}{(r^2+\varepsilon)^{1/2}} + \frac{x_ix_j}{(r^2+\varepsilon)^{3/2}}
~\rightarrow~ {\rm P.V.}\left(\frac{\delta_{ij}}{r} + \frac{x_ix_j}{r^3}\right)
\quad\text{for}\quad\varepsilon\to 0^+ \tag{A}$$
in $C^{\infty}(\mathbb{R}^3)$, in the sense of generalized functions, then the derivatives are well-defined:
$$\frac{\partial T_{ij}}{\partial x^k}
~=~\frac{\delta_{ik}x_j +\delta_{jk}x_i- \delta_{ij}x_k}{(r^2+\varepsilon)^{3/2}} - 3\frac{x_ix_jx_k}{(r^2+\varepsilon)^{5/2}}, \tag{B}$$
$$ \begin{align}\nabla^2T_{ij}&~=~-\frac{\delta_{ij}}{(r^2+\varepsilon)^{3/2}} + 3\frac{\delta_{ij}r^2-7x_ix_j}{(r^2+\varepsilon)^{5/2}} +15\frac{x_ix_jr^2}{(r^2+\varepsilon)^{7/2}} \cr
&~=~\delta_{ij}\left(\frac{2}{(r^2+\varepsilon)^{3/2}}
-\color{red}{\frac{3\varepsilon}{(r^2+\varepsilon)^{5/2}}}\right)
 - 3x_ix_j\left(\frac{2}{(r^2+\varepsilon)^{5/2}}
+\color{red}{\frac{5\varepsilon}{(r^2+\varepsilon)^{7/2}}}\right)\cr
&~\rightarrow~ \delta_{ij}\left({\rm P.V.}\frac{2}{r^3} - \color{red}{8\pi\delta^3({\bf r})}\right) - {\rm P.V.} \frac{6x_ix_j}{r^5} \quad\text{for}\quad\varepsilon\to 0^+ .\end{align} \tag{C}$$
In eq. (C) the 1st & 3rd terms reproduce the terms in eq. (5.118). The 2nd & 4th term are indeed 3D Dirac delta distributions (of same size). Interestingly, there are no Dirac delta contributions in the off-diagonal sectors $i\!\neq\!j$. $\Box$
--
$^1$ Let us for simplicity drop an overall trivial multiplicative factor $\frac{1}{8\pi\eta_0}$. Here the principal value distributions 
$${\rm P.V.} \frac{1}{r^p}~:=~\lim_{\varepsilon\to 0^+}\frac{1}{(r^2+\varepsilon)^{p/2}}, \qquad p~\leq~3,\tag{D}$$
$${\rm P.V.} \frac{x_ix_j}{r^p}~:=~\lim_{\varepsilon\to 0^+} \frac{x_ix_j}{(r^2+\varepsilon)^{p/2}}, \qquad p~\leq~5,\tag{E}$$
are well-defined for test functions  $\in C^{\infty}_c(\mathbb{R}^3)$ that vanish at $r=0$ for the observed powers $p$. See also this related math.SE post.
