Proof of $z^{-1 + \epsilon} = \frac{1}{\epsilon} \delta(z) + \sum_{k = 0}^\infty \frac{\epsilon^k}{k!} \left( \frac{\ln z}{z} \right)_+$ I came across this formula in some paper $$z^{-1 + \epsilon} = \frac{1}{\epsilon} \delta(z) + \sum_{k = 0}^\infty \frac{\epsilon^k}{k!} \left( \frac{\ln z}{z} \right)_+$$ Where the plus indicates the plus distribution and $\delta$ is the usual delta function.
There is no proof given nor is there any reference to an other paper. Does anybody have an idea how to prove this?
 A: For $\varepsilon >0$, we define the distribution $\psi_{\varepsilon}(z)=\left(z^{\varepsilon - 1}\right)_+$ as
$$\langle \psi_\varepsilon , \phi\rangle = \int_0^\infty z^{\varepsilon - 1}\phi(z)\,dz$$
Integrating by parts with $u=\phi$ and $v=\frac{z^\varepsilon}{\varepsilon}$ reveals
$$\begin{align}
\langle \psi_\varepsilon , \phi\rangle &= -\frac1\varepsilon\int_0^\infty z^{\varepsilon }\phi'(z)\,dz\\\\
&=-\frac1\varepsilon\int_0^\infty e^{\varepsilon \log(z)}\phi'(z)\,dz\\\\
&=\frac{\phi(0)}{\varepsilon}-\sum_{k=0}^\infty \frac{\varepsilon^{k}}{(k+1)!}\int_0^\infty \log^{k+1}(z) \phi'(z)\,dz\\\\
&=\frac{\phi(0)}{\varepsilon}+\sum_{k=0}^\infty \frac{\varepsilon^{k}}{k!}\int_0^\infty \left(-\frac{\log^{k+1}(z)}{k+1}\right) \phi'(z)\,dz\\\\
\end{align}$$
Therefore, we can write in distribution that
$$\left(z^{\varepsilon - 1}\right)_+=\frac1\varepsilon \delta(z)+\sum_{k=0}^\infty \frac{\varepsilon^k}{k!}\left(\frac{\log^k(z)}{z}\right)_+$$
where the distribution $\ell_k(z)=\left(\frac{\log^k(z)}{z}\right)_+$ is defined as
$$\langle \ell_k,\phi\rangle = \int_0^\infty \left(-\frac{\log^{k+1}(z)}{k+1}\right)\phi'(z)\,dz$$
as was to be shown!
