Question Number 190779 by 07049753053 last updated on 11/Apr/23
$$\boldsymbol{\mathrm{R}{e}}\left(\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{{a}}}\right)^{\boldsymbol{{a}}−\mathrm{1}} \\ $$
Answered by Frix last updated on 11/Apr/23
$${f}\left({a}\right)=\left(\frac{\mathrm{1}}{\mathrm{1}−{a}}\right)^{{a}−\mathrm{1}} \in\mathbb{R}\forall{a}<\mathrm{1} \\ $$$$\underset{{a}\rightarrow\mathrm{1}} {\mathrm{lim}}\:{f}\left({a}\right)\:=\mathrm{1} \\ $$$$\forall{a}>\mathrm{1}:\:{f}\left({a}\right)=\frac{\mathrm{1}}{\left({a}−\mathrm{1}\right)^{{a}} }\left(\left(\mathrm{1}−{a}\right)\mathrm{cos}\:\pi{a}\:+\left(\mathrm{1}−{a}\right)\mathrm{i}\:\mathrm{sin}\:\pi{a}\right) \\ $$$$\:\:\:\:\:\forall{a}\in\mathbb{N}^{\bigstar} :\:{f}\left({a}\right)=−\left(−\mathrm{1}\right)^{{a}} \left({a}−\mathrm{1}\right)^{\mathrm{1}−{a}} \in\mathbb{R} \\ $$$$\:\:\:\:\:\forall{a}>\mathrm{1}\wedge{a}\notin\mathbb{N}:\:\mathrm{Re}\:\left({f}\left({a}\right)\right)=−\left({a}−\mathrm{1}\right)^{\mathrm{1}−{a}} \mathrm{cos}\:\pi{a} \\ $$
Commented by 07049753053 last updated on 12/Apr/23
$$\boldsymbol{\mathrm{thanks}} \\ $$