${\displaystyle g(v)=\int \delta (yf(z)-z+x)g(z)(1-yf'(z))\,dz}$

Escribiendo la delta de Dirac en forma integral, se tiene:

${\displaystyle {\begin{aligned}g(v)&=\int \int \exp(ik[yf(z)-z+x])g(z)(1-yf'(z))\,{\frac {dk}{2\pi }}\,dz\\[10pt]&=\sum _{n=0}^{\infty }\int \int {\frac {(ikyf(z))^{n}}{n!}}g(z)(1-yf'(z))e^{ik(x-z)}\,{\frac {dk}{2\pi }}\,dz\\[10pt]&=\sum _{n=0}^{\infty }\left({\frac {\partial }{\partial x}}\right)^{n}\int \int {\frac {(yf(z))^{n}}{n!}}g(z)(1-yf'(z))e^{ik(x-z)}\,{\frac {dk}{2\pi }}\,dz\end{aligned}}}$

La integral sobre ${\displaystyle k}$ da ${\displaystyle \delta (x-z)}$ (pues es otra vez la representación integral de la delta de Dirac), por lo que se puede resolver también la integral sobre ${\displaystyle z}$ obteniendo la serie

${\displaystyle {\begin{aligned}g(v)&=\sum _{n=0}^{\infty }\left({\frac {\partial }{\partial x}}\right)^{n}\left[{\frac {(yf(x))^{n}}{n!}}g(x)(1-yf'(x))\right]\\[10pt]&=\sum _{n=0}^{\infty }\left({\frac {\partial }{\partial x}}\right)^{n}\left[{\frac {y^{n}f(x)^{n}g(x)}{n!}}-{\frac {y^{n+1}}{(n+1)!}}\left\{(g(x)f(x)^{n+1})'-g'(x)f(x)^{n+1}\right\}\right]\end{aligned}}}$

Reordenando la serie, se obtiene el resultado buscado:

${\displaystyle g(v)=g(x)+\sum _{k=1}^{\infty }{\frac {y^{k}}{k!}}\left({\frac {\partial }{\partial x}}\right)^{k-1}\left(f(x)^{k}g'(x)\right)}$