The definition of the moment of inertia tensor $I_{ij}$ used in the proof, is: $I_{ij} = \int_V \rho (x_k x_k \delta_{ij} - x_i x_j) dV$

This integral formula defines the components of the moment of inertia tensor for a rigid body rotating about an axis passing through the origin, where:

• $I_{ij}$ are the components of the moment of inertia tensor (a $3 \times 3$ matrix).
• $\int_V$ denotes the integral over the volume of the body.
• $\rho$ is the mass density of the body (mass per unit volume).
• $x_i$ are the Cartesian coordinates of a mass element within the body.
• $x_k x_k = \sum_{k=1}^3 x_k^2$ is the square of the distance from the origin, $r^2$.
• $\delta_{ij}$ is the Kronecker delta, which is 1 if $i=j$ and 0 if $i \neq j$.
• $dV$ is the volume element.

In component form:

• The diagonal components ($i=j$) represent the moment of inertia about the $i$-axis:

  $I_{ii} = \int_V \rho (x_k x_k \delta_{ii} - x_i x_i) dV = \int_V \rho (r^2 - x_i^2) dV$

• The off-diagonal components ($i \neq j$) are the products of inertia:

  $I_{ij} = \int_V \rho (x_k x_k \cdot 0 - x_i x_j) dV = - \int_V \rho x_i x_j dV$

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