find-laurant-series-at-f-z-1-lnz-

Question Number 145052 by tabata last updated on 01/Jul/21

f i n d l a u r a n t s e r i e s a t f (z) = \frac{1}{l n z} ? ? ? ?

Commented by tabata last updated on 01/Jul/21

? ? ? ? ?

Commented by Olaf_Thorendsen last updated on 03/Jul/21

li (z) = \int \frac{d z}{\ln (z)}

Expansion about z = 1 :

li (z) = \frac{1}{2} (\ln (z - 1) - \ln (\frac{1}{z - 1}))

+ γ + \underset{k = 0}{\sum^{\infty}} \frac{{(- 1)}^{k}}{(k + 1)!} {(1 - z)}^{k + 1} \underset{j = 1}{\sum^{k + 1}} \frac{B_{j} S_{k}^{(j - 1)}}{j}

B_{n} : n t h B e r n o u l l i n u m b e r

S_{n}^{(m)} : s i g n e d S t i r l i n g n u m b e r

o f t h e f i r s t k i n d

\Rightarrow \frac{1}{\ln (z)} = \frac{1}{z - 1} - \underset{k = 0}{\sum^{\infty}} \underset{j = 1}{\sum^{k + 1}} \frac{B_{j} S_{k}^{(j - 1)}}{k! j} {(z - 1)}^{k}