In a paper I am reading, the authors claime that by invoking the power law representation of the $\delta$ distribution this gives
$$ \delta(k) = (1/4\pi)\lim_{\tau\rightarrow 0} \tau k^{-3 + \tau} $$
in three dimensional space.
From a naive point of view it looks as if the number 3 in the above equation corresponds to the number of dimensions such that for D dimensional space one would have
$$ \delta(k) = (1/4\pi)\lim_{\tau\rightarrow 0} \tau k^{-D + \tau} $$
I am not familiar with this representation of the $\delta$ distribution, so can somebody explain to me how this representation can be derived generally for D dimensional space?
Indeed, in each dimension $D\geqslant1$, $$ \delta(k)=\lim_{\tau\to0}\frac1{\sigma_D}\tau\|k\|^{-D+\tau}, $$ where $\sigma_D$ denotes the surface of the unit sphere $S^{D-1}$ in dimension $D$. To check this, recall that, if $k=ru$ with $r\gt0$ and $u$ in $S^{D-1}$, the element of volume is $$ \mathrm dk=r^{D-1}\mathrm dr\,\mathrm ds_D(u), $$ Thus, for every $R\gt0$, $$ \iint_{B(0,R)}\frac1{\sigma_D}\tau\|k\|^{-D+\tau}\mathrm dk=\int_0^R\tau r^{-D+\tau}(r^{D-1}\mathrm dr)\int_{S^{D-1}}\frac1{\sigma_D}\mathrm ds_D(u)=R^\tau, $$ hence, when $\tau\to0$, $$ \iint_{B(0,R)}\frac1{\sigma_D}\tau\|k\|^{-D+\tau}\mathrm dk\to1. $$