Several derivatives of adjoined distributions (Q2781723)

Since \((x\pm i0)^\lambda= \exp\{\lambda\ln (x\pm i0)\}\) are entire functions of \(\lambda\), \((x\pm i0)^\lambda\ln^k(x\pm i0)\), given by \(k\)-times differentiations, are also entire functions. The author proves the two representations:NEWLINENEWLINENEWLINE\((x\pm i0)^\lambda\ln^k(x\pm i0)= x^\lambda_+\ln^k x_++ e^{\pm i\lambda\pi}\sum_{j= 0\sim k}\{{_kC_j}(\pm i\pi)^{k- j}x^\lambda_-\ln^j x_-\}\), andNEWLINENEWLINENEWLINE\((x\pm i0)^{-n} \ln^k(x\pm i0)= x^{-n}_+ \ln^k x_++ (-1)^n \sum_{j= 0\sim k}\{{_kC_j}(\pm i\pi)^{k- j}\cdot x^{-n}_- \ln^j x_-\}+ (-1)^n (\pm i\pi)^{k+1}\) \(\delta^{(n-1)}(x)/\{(k+ 1)(n- 1)!\}\),NEWLINENEWLINENEWLINEwhere \(\lambda\in \mathbb{C}\setminus I_-\), \(I_-= \{\lambda i\); \(\lambda<0\}\).NEWLINENEWLINENEWLINEHe uses the Taylor-series (resp. Laurent-series) expansions of \((x\pm i0)^\lambda\), \(x^\lambda_\pm\), and \(e^{\pm i\lambda\pi}\) in powers of \((\lambda- \lambda_0)\) (resp. of \((\lambda+ n)\)) for the proof. Next, he proves the formulae of derivatives \((d/dx)(\ln^k x_+)= kx^{-1}_+ \ln^{k-1} x_+\), etc. by using testing functions \(\phi\in{\mathcal D}\), and derives the two formulaeNEWLINENEWLINENEWLINE\((d/dx)\{(x\pm i0)^\lambda \ln^k(x\pm i0)\}= \lambda(x\pm i0)^{\lambda- 1}\ln^k(x\pm i0)+ k(x\pm i0)^{\lambda- 1}\ln^{k-1}(x\pm i0)\), andNEWLINENEWLINENEWLINE\((d/dx)\{(x\pm i0)^{-n} \ln^k(x\pm i0)\}= -n(x\pm i0)^{-n-1} \ln^k(x\pm i0)+ k(x\pm i0)^{-n-1} \ln^{k-1}(x\pm i0)\).