How do I calculate the EM fields of a massless charged particle?

In the referenced paper we read

It is straightforward to establish (for example by Fourier transforming with respect to $z$) that begin{equation} boxed{:: lim_{upsilonboldsymbol{rightarrow} 1}dfrac{1boldsymbol{-}upsilon^2}{left[left(zboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,3/2}}boldsymbol{=}dfrac{2}{r^2_{boldsymbol{perp}}},deltaleft(tboldsymbol{-}zright)::vphantom{dfrac{dfrac{a}{b}}{dfrac{a}{b}}}} tag{4 of paper}label{4 of paper} end{equation}

Instead of Fourier transforming I'll try to follow a different way.

So, consider a real function $;f(z);$ of the real variable $;zinmathbb{R};$ for which begin{align} f(z)boldsymbol{=}0 quad & text{for any} quad zboldsymbol{ne} z_{0} quad textbf{and} tag{01a}label{01a} intlimits_{boldsymbol{z_{0}-varepsilon}}^{boldsymbol{z_{0}+varepsilon}}!!!f(z)mathrm dzboldsymbol{=}1quad & text{for any} quad boldsymbol{varepsilon} boldsymbol{>}0 tag{01b}label{01b} end{align} Under these conditions it seems that this function is not well-defined at $;z_{0}$, may be because of a singularity at this point. But we have good reasons to $^{primeprime}$believe$^{primeprime}$ that
begin{equation} f(z)boldsymbol{equiv}deltaleft(zboldsymbol{-}z_{0}right) tag{02}label{02} end{equation} since equations eqref{01a},eqref{01b} remind us the defining properties of Dirac delta function on the real axis $;mathbb{R}$.

In our case we have the function begin{equation} f(z)boldsymbol{=}lim_{upsilonboldsymbol{rightarrow} 1}dfrac{1boldsymbol{-}upsilon^2}{left[left(zboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,3/2}} tag{03}label{03} end{equation} This function fulfills the condition eqref{01a} with $z_{0}=t$ begin{equation} f(z)boldsymbol{=}lim_{upsilonboldsymbol{rightarrow} 1}dfrac{1boldsymbol{-}upsilon^2}{left[left(zboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,3/2}}boldsymbol{=}0 qquadtext{for any} quad zboldsymbol{ne} t tag{04}label{04} end{equation} But this function is not well-defined at $;zboldsymbol{=}t$ since given that $0<upsilon<1$ begin{align} f(t) & boldsymbol{=}lim_{upsilonboldsymbol{rightarrow} 1}dfrac{1boldsymbol{-}upsilon^2}{left[left(tboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,3/2}} nonumber & boldsymbol{=}lim_{upsilonboldsymbol{rightarrow} 1}dfrac{1boldsymbol{+}upsilon}{left(1boldsymbol{-}upsilonright)^{,1/2}left[left(1boldsymbol{-}upsilonright)t^2boldsymbol{+}left(1boldsymbol{+}upsilonright)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,3/2}}boldsymbol{=}boldsymbol{+}infty tag{05}label{05} end{align} We'll see now if our function fulfills a condition similar to eqref{01b}. For $boldsymbol{varepsilon} boldsymbol{>}0$ we have begin{align} intlimits_{boldsymbol{t-varepsilon}}^{boldsymbol{t+varepsilon}}!!!f(z)mathrm dz &boldsymbol{=}intlimits_{boldsymbol{t-varepsilon}}^{boldsymbol{t+varepsilon}}!!!lim_{upsilonboldsymbol{rightarrow} 1}dfrac{1boldsymbol{-}upsilon^2}{left[left(zboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,3/2}}mathrm dz nonumber &boldsymbol{=}lim_{upsilonboldsymbol{rightarrow} 1}intlimits_{boldsymbol{t-varepsilon}}^{boldsymbol{t+varepsilon}}!!!dfrac{1boldsymbol{-}upsilon^2}{left[left(zboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,3/2}}mathrm dz nonumber &boldsymbol{=}lim_{upsilonboldsymbol{rightarrow} 1}left[dfrac{zboldsymbol{-}upsilon,t}{r^2_{boldsymbol{perp}}left[left(zboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,1/2}}vphantom{dfrac{dfrac{a}{b}}{dfrac{a}{b}}}right]_{zboldsymbol{=}boldsymbol{t-varepsilon}}^{zboldsymbol{=}boldsymbol{t+varepsilon}} nonumber &boldsymbol{=}dfrac{1}{r^2_{boldsymbol{perp}}}left[lim_{upsilonboldsymbol{rightarrow} 1}dfrac{zboldsymbol{-}upsilon,t}{left[left(zboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,1/2}}vphantom{dfrac{dfrac{a}{b}}{dfrac{a}{b}}}right]_{zboldsymbol{=}boldsymbol{t-varepsilon}}^{zboldsymbol{=}boldsymbol{t+varepsilon}} nonumber &boldsymbol{=}dfrac{1}{r^2_{boldsymbol{perp}}}left[dfrac{zboldsymbol{-}t}{boldsymbol{vert}zboldsymbol{-}tboldsymbol{vert}}right]_{zboldsymbol{=}boldsymbol{t-varepsilon}}^{zboldsymbol{=}boldsymbol{t+varepsilon}} boldsymbol{=}dfrac{1}{r^2_{boldsymbol{perp}}}left(dfrac{varepsilon}{boldsymbol{vert}varepsilonboldsymbol{vert}}boldsymbol{-}dfrac{boldsymbol{-}varepsilon}{boldsymbol{vert}boldsymbol{-}varepsilonboldsymbol{vert}}right) nonumber &boldsymbol{=}dfrac{1}{r^2_{boldsymbol{perp}}}dfrac{2varepsilon}{boldsymbol{vert}varepsilonboldsymbol{vert}}boldsymbol{=}dfrac{2}{r^2_{boldsymbol{perp}}} tag{06}label{06} end{align} In equation eqref{06} the second equality is valid under the assumption that the limit and integral operators commute, while the third equality is due to the following indefinite integral ^{begin{equation} intdfrac{1}{left[left(zboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,3/2}}mathrm dzboldsymbol{=}dfrac{zboldsymbol{-}upsilon,t}{left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}left[left(zboldsymbol{-}upsilon,tright)^2boldsymbol{+}left(1boldsymbol{-}upsilon^2right)r^2_{boldsymbol{perp}}vphantom{tfrac{a}{b}}right]^{,1/2}}boldsymbol{+}texttt{const.} tag{07}label{07} end{equation}}

By equations eqref{04} and eqref{06} we have proved equation eqref{4 of paper}.