Question Number 196023 by pticantor last updated on 16/Aug/23
$${find}\:{the}\:{domain}\:{of}\:{definition}\:{of}\:{this} \\ $$$$\left.{function}\:{for}\:{t}\in\right]\mathrm{0};\mathrm{1}\left[\right. \\ $$$$\:\:\:\:\:\boldsymbol{\rho}\left({x}\right)=\int_{{x}} ^{\mathrm{2}{x}} \frac{\mathrm{1}}{{lnt}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{ptiCantor} \\ $$
Answered by sniper237 last updated on 16/Aug/23
$$\left.{t}\rightarrow\frac{\mathrm{1}}{{lnt}}\:\:{is}\:{continue}\:{on}\:\:\:\right]\mathrm{0};\mathrm{1}\left[\:,\:{and}\:\right]\mathrm{1};+\infty\left[\right. \\ $$$${p}\left({x}\right)\:{exist}\:\:{if}\:\:\:\mathrm{2}{x}\:,\:{x}\:{are}\:{in}\:{the}\:{same}\:{interval} \\ $$$$\:\:\Rightarrow\:\:\mathrm{0}<\:{x}<\mathrm{1}/\mathrm{2}\:\:\:{or}\:\:\:\mathrm{1}<\:{x}\: \\ $$$$\: \\ $$
Commented by York12 last updated on 16/Aug/23
$${bro}\:{please}\:{look}\:{at}\:{the}\:{above}\:{problem} \\ $$
Commented by York12 last updated on 16/Aug/23
$$\mathrm{196206} \\ $$
Commented by York12 last updated on 16/Aug/23
$${sorry}\:{I}\:{meant}\:\mathrm{196204} \\ $$