Question Number 145052 by tabata last updated on 01/Jul/21
$${find}\:{laurant}\:{series}\:{at}\:{f}\left({z}\right)=\frac{\mathrm{1}}{{lnz}}???? \\ $$
Commented by tabata last updated on 01/Jul/21
$$????? \\ $$
Commented by Olaf_Thorendsen last updated on 03/Jul/21
$$\mathrm{li}\left({z}\right)\:=\:\int\frac{{dz}}{\mathrm{ln}\left({z}\right)} \\ $$$$\mathrm{Expansion}\:\mathrm{about}\:{z}\:=\:\mathrm{1}\:: \\ $$$$\mathrm{li}\left({z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\left({z}−\mathrm{1}\right)−\mathrm{ln}\left(\frac{\mathrm{1}}{{z}−\mathrm{1}}\right)\right) \\ $$$$+\gamma+\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)!}\left(\mathrm{1}−{z}\right)^{{k}+\mathrm{1}} \underset{{j}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\sum}}\frac{{B}_{{j}} {S}_{{k}} ^{\left({j}−\mathrm{1}\right)} }{{j}} \\ $$$${B}_{{n}} \::\:{nth}\:{Bernoulli}\:{number} \\ $$$${S}_{{n}} ^{\left({m}\right)} \::\:{signed}\:{Stirling}\:{number} \\ $$$${of}\:{the}\:{first}\:{kind} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{ln}\left({z}\right)}\:=\:\frac{\mathrm{1}}{{z}−\mathrm{1}}−\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{k}+\mathrm{1}} {\sum}}\frac{{B}_{{j}} {S}_{{k}} ^{\left({j}−\mathrm{1}\right)} }{{k}!{j}}\left({z}−\mathrm{1}\right)^{{k}} \\ $$