Li-z-2-z-1-ln-t-dt-Li-a-ib-R-2-a-ib-1-ln-t-dt-iI-2-a-ib-1-ln-t-dt-Can-you-explain-me-how-get-the-formula-for-R-Li-z-and-I-Li-z-

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

$$\mathrm{Li}\left({z}\right)=\int_{\mathrm{2}} ^{{z}} \frac{\mathrm{1}}{\mathrm{ln}\:{t}}\mathrm{d}{t} \\ $$$$\mathrm{Li}\left({a}+{ib}\right)=\mathfrak{R}\left(\int_{\mathrm{2}} ^{{a}+{ib}} \frac{\mathrm{1}}{\mathrm{ln}\:{t}}\mathrm{d}{t}\right)+{i}\mathfrak{I}\left(\int_{\mathrm{2}} ^{{a}+{ib}} \frac{\mathrm{1}}{\mathrm{ln}\:{t}}\mathrm{d}{t}\right) \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{me} \\ $$$$\mathrm{how}\:\mathrm{get}\:\mathrm{the}\:\mathrm{formula} \\ $$$$\mathrm{for}\:\mathfrak{R}\left(\mathrm{Li}\left({z}\right)\right)\:\mathrm{and} \\ $$$$\mathfrak{I}\left(\mathrm{Li}\left({z}\right)\right)? \\ $$

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

L_i (z) =∫_2 ^z (dt/(lnt)) we take z =a+ib ⇒L_i (a+ib) =∫_2 ^(a+ib) (dt/(lnt)) this integral is complex so ∃ α and β from R /L_i (a+ib) =α +iβ α =Re(∫_2 ^(a+ib) (dt/(lnt))) and β =Im(∫_2 ^(a+ib) (dt/(lnt))) the problem here is how to find α and β...!

$$\mathrm{L}_{\mathrm{i}} \left(\mathrm{z}\right)\:=\int_{\mathrm{2}} ^{\mathrm{z}} \:\frac{\mathrm{dt}}{\mathrm{lnt}}\:\:\:\mathrm{we}\:\mathrm{take}\:\mathrm{z}\:=\mathrm{a}+\mathrm{ib}\:\Rightarrow\mathrm{L}_{\mathrm{i}} \left(\mathrm{a}+\mathrm{ib}\right)\:=\int_{\mathrm{2}} ^{\mathrm{a}+\mathrm{ib}} \:\frac{\mathrm{dt}}{\mathrm{lnt}} \\ $$$$\mathrm{this}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{complex}\:\:\mathrm{so}\:\exists\:\alpha\:\mathrm{and}\:\beta\:\mathrm{from}\:\mathrm{R}\:/\mathrm{L}_{\mathrm{i}} \left(\mathrm{a}+\mathrm{ib}\right)\:=\alpha\:+\mathrm{i}\beta \\ $$$$\alpha\:=\mathrm{Re}\left(\int_{\mathrm{2}} ^{\mathrm{a}+\mathrm{ib}} \:\frac{\mathrm{dt}}{\mathrm{lnt}}\right)\:\mathrm{and}\:\beta\:=\mathrm{Im}\left(\int_{\mathrm{2}} ^{\mathrm{a}+\mathrm{ib}} \:\frac{\mathrm{dt}}{\mathrm{lnt}}\right) \\ $$$$\mathrm{the}\:\mathrm{problem}\:\mathrm{here}\:\mathrm{is}\:\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\:\alpha\:\mathrm{and}\:\beta…! \\ $$

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

Exatly, but I don′t know how can we do that :/

$$\mathrm{Exatly},\:\mathrm{but}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know} \\ $$$$\mathrm{how}\:\mathrm{can}\:\mathrm{we}\:\mathrm{do}\:\mathrm{that}\::/ \\ $$

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