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Question Number 103124 by Faetma last updated on 12/Jul/20

$$\mathrm{Li}\left(z\right)={\int }_2^z\frac{1}{\ln t}\mathrm{d}t$$$$\mathrm{Li}(a+ib)=\mathfrak{R}\left({\int }_2^{a+ib}\frac{1}{\ln t}\mathrm{d}t\right)+i\mathfrak{I}\left({\int }_2^{a+ib}\frac{1}{\ln t}\mathrm{d}t\right)$$$$\mathrm{Can}\mathrm{you}\mathrm{explain}\mathrm{me}$$$$\mathrm{how}\mathrm{get}\mathrm{the}\mathrm{formula}$$$$\mathrm{for}\mathfrak{R}\left(\mathrm{Li}\left(z\right)\right)\mathrm{and}$$$$\mathfrak{I}\left(\mathrm{Li}\left(z\right)\right)?$$

Answered by mathmax by abdo last updated on 13/Jul/20

$${\mathrm{L}}_{\mathrm{i}}\left(\mathrm{z}\right)={\int }_2^{\mathrm{z}}\frac{\mathrm{dt}}{\mathrm{lnt}}\mathrm{we}\mathrm{take}\mathrm{z}=\mathrm{a}+\mathrm{ib}\Rightarrow {\mathrm{L}}_{\mathrm{i}}(\mathrm{a}+\mathrm{ib})={\int }_2^{\mathrm{a}+\mathrm{ib}}\frac{\mathrm{dt}}{\mathrm{lnt}}$$$$\mathrm{this}\mathrm{integral}\mathrm{is}\mathrm{complex}\mathrm{so}\mathrm{\exists }\alpha \mathrm{and}\beta \mathrm{from}\mathrm{R}/{\mathrm{L}}_{\mathrm{i}}(\mathrm{a}+\mathrm{ib})=\alpha +\mathrm{i}\beta$$$$\alpha =\mathrm{Re}\left({\int }_2^{\mathrm{a}+\mathrm{ib}}\frac{\mathrm{dt}}{\mathrm{lnt}}\right)\mathrm{and}\beta =\mathrm{Im}\left({\int }_2^{\mathrm{a}+\mathrm{ib}}\frac{\mathrm{dt}}{\mathrm{lnt}}\right)$$$$\mathrm{the}\mathrm{problem}\mathrm{here}\mathrm{is}\mathrm{how}\mathrm{to}\mathrm{find}\alpha \mathrm{and}\beta ...!$$

Commented by mathmax by abdo last updated on 13/Jul/20

$$\mathrm{your}\mathrm{cut}\mathrm{is}\mathrm{crying}\mathrm{why}...?$$

Commented by Faetma last updated on 13/Jul/20

$$\mathrm{Exatly},\mathrm{but}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}$$$$\mathrm{how}\mathrm{can}\mathrm{we}\mathrm{do}\mathrm{that}:/$$

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