Inertia tensor integral $$I_{xx}=\sum{m_\alpha}(y_\alpha^2+z_\alpha^2)$$
$$ I_{xy}=-\sum m_\alpha x_\alpha y_\alpha $$
How can these two definitions of the inertia tensor be combined into one definition to get $$ I_{ij}=\int \rho (r^2\delta_{ij}-r_i r_j)dV? $$
 A: 
$$I_{xx}=\sum{m_\alpha}(y_\alpha^2+z_\alpha^2)$$
  $$ I_{xy}=-\sum m_\alpha x_\alpha y_\alpha $$
  How can these two definitions of the inertia tensor be combined into one?

Those expressions only cover two of the six independent elements of the inertia tensor of a collection of point masses. The expression for $I_{xy}$ obviously generalizes to
$$I_{ij}=-\sum m_\alpha r_{\alpha,i}r_{\alpha,j}\quad(i \ne j) \tag{1}$$
A way to generalize the first expression to describe the three diagonal elements of the inertia tensor is to first recognize that $y_\alpha^2 + z_\alpha^2$ is equal to $r_\alpha^2 - x_\alpha^2$, leading to
$$I_{xx}=\sum{m_\alpha}(r_\alpha^2-x_\alpha^2)$$
This generalizes to
$$I_{ii}=\sum{m_\alpha}(r_\alpha^2-r_{\alpha,i}^2)\tag{2}$$
Combining expressions (1) and (2) yields
$$
I_{ij} =
  \begin{cases}
    \sum m_\alpha (r_\alpha^2-r_{\alpha,i}r_{\alpha,j}) & j=i \\
    \sum m_\alpha (\phantom{r_\alpha^2}-r_{\alpha,i}r_{\alpha,j}) & j \ne i
  \end{cases}\tag{3}
$$
Equation (3) can be succinctly written using the Kronecker delta as
$$I_{ij}=\sum{m_\alpha}(r_\alpha^2\delta_{ij}-r_{\alpha,i}r_{\alpha,j})$$
For a mass distribution, the above becomes, with a bit of abuse of notation,
$$I_{ij}=\int(r^2\delta_{ij}-r_i r_j)\,dm$$
Recognizing that $dm = \rho dV$, this becomes
$$I_{ij}=\int \rho (r^2\delta_{ij}-r_i r_j)\,dV$$
