QED integral is zero in dimensional regularization

Question

Why is this integral zero in dimensional regularization? $$ \int\frac{d^Dk}{(2\pi)^D}\frac{1}{(k^2)^n}. $$

Answer 1

1. If $n\neq D/2$ then OP's integral is dimensionful (i.e. it has non-zero mass dimension $D-2n\neq 0$ in natural units), but it doesn't depend on any dimensionful parameters, so the only possible consistent answer of the regularization is zero.

2. If $n=D/2$ then OP's integral is logarithmically divergent, and in fact non-zero in dimensional regularization.

Answer 2

See around (8.33) of Regularisation. This is known as Veltman’s formula. See also post and post.

$$ \int d^n k \frac{1}{k^a} = \lim_{m^2 \to 0} \int d^n k \frac{1}{\left( k^2 - m^2 \right)^{a/2}} \propto \lim_{m^2 \to 0} m^{n-a} $$

So integral is zero for $n>a$.