# Is there a completion of Laurent series w.r.t. integration?

Given a Laurent series $$f(z) = \sum_{k=-\infty}^\infty a_k(z-z_0)^k,$$ its derivative is obviously also a Laurent series. However, upon integration, the $\frac1{z-z_0}$ term introduces a $\ln(z-z_0)$ term and therewith an essential singularity turning the integral of a Laurent series into something that generally is not a Laurent series again. So one could define a new kind of series including that $\ln$ term, and upon integration obtain another new term due to $\int\ln z\,dz = z(\ln z-1)$, or more generally, $$\int z^k\ln z\,dz = \frac{z^{k+1}[(k+1)\ln z -1]}{(k+1)^2}$$ for $k\neq-1$.

So the series $$f(z) := \sum_{k=-\infty}^\infty [a_k+b_k\ln(z-z_0)](z-z_0)^k$$ with $b_{-1}=0$ is complete under both derivation and integration at the expense of introducing a branch cut and an essential singularity at $z_0$. But does such a series definition make sense? How would one obtain the coefficients $b_k$ now?

For $b_{-1}\neq0$, one obtains $\int\frac{\ln z}z\,dz = \frac12\ln^2 z$, the inclusion of which requires the inclusion of $\int\ln^2z\,dz = z(\ln^2z -2\ln z+2)$ and therewith of $z^m\ln^2z$ for $m\neq-1$. And should the extended series include a $\frac{\ln^2z}z$ term, its integration would require a $\ln^3z$ term and so on, so in an even more general sense we'd have $$f(z) := \sum_{k=-\infty}^\infty \sum_{l=0}^\infty c_{kl} \ln^l(z-z_0)\cdot (z-z_0)^k.$$ Now you could go on and include $l<0$ terms, but their integrals would require the inclusion of the logarithmic integral $\operatorname{li}(z-z_0)$, and integrating that would include $\operatorname{Ei}(z-z_0)$ and hypergeometric functions etc. - so, does this ever stop at a (up to powers) finite set of base functions?

• For $k = -1$ the integral $\int \frac{\log z}z\,dz = \int z^k\,\log z\, dz$ is not elementary. Same problem as before ... – martini Sep 30 '13 at 8:50
• @martini You're right, which leads to including $\ln^2 z$, which leads to including $\int\ln^2z/z\propto\ln^3z$ and so on, so all powers of $\ln$ would need to be included as well... So the $[a_k+b_k\ln(z-z_0)]$ actually becomes a power series of $\ln(z-z_0)$, i.e. $$f(z) := \sum_{k=-\infty}^\infty \sum_{l=0}^\infty c_{kl}\ln^l(z-z_0)\cdot(z-z_0)^k$$ – Tobias Kienzler Sep 30 '13 at 8:59