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

$$findlaurantseriesatf\left(z\right)=\frac{1}{lnz}????$$

Commented by tabata last updated on 01/Jul/21

$$?????$$

Commented by Olaf_Thorendsen last updated on 03/Jul/21

$$\mathrm{li}\left(z\right)=\int \frac{dz}{\ln \left(z\right)}$$$$\mathrm{Expansion}\mathrm{about}z=1:$$$$\mathrm{li}\left(z\right)=\frac{1}{2}(\ln (z-1)-\ln \left(\frac{1}{z-1}\right))$$$$+\gamma +\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{(k+1)!}{(1-z)}^{k+1}\underset{j=1}{\sum ^{k+1}}\frac{B_j{S}_k^{(j-1)}}{j}$$$$B_n:nthBernoullinumber$$$$S_n^{\left(m\right)}:signedStirlingnumber$$$$ofthefirstkind$$$$\Rightarrow \frac{1}{\ln \left(z\right)}=\frac{1}{z-1}-\underset{k=0}{\sum ^{\mathrm{\infty }}}\underset{j=1}{\sum ^{k+1}}\frac{B_j{S}_k^{(j-1)}}{k!j}{(z-1)}^k$$

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