 Finite tensor-integral remainder in dimensional regularization

Include the measure factor $(2\pi {)}^{-D}$ and a conventional renormalization scale factor in the displayed integral. The Lorentz invariant replacement $k_{\mu }{k}_{\nu }\mapsto g_{\mu \nu }{k}^2/D$ must be performed before taking $D\to 4$. For $D=4-\epsilon$, its scalar integral has pole $i/(8{\pi }^2\epsilon )$, while $4/D-1=\epsilon /4+O({\epsilon }^2)$. Their product has the finite value shown. Setting $D=4$ before integration discards this value and can spoil a Ward identity, even in a final ultraviolet-finite scattering amplitude. This is an elementary example of a vanishing regulator factor multiplying a divergent integral.