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Question Number 142443 by Hassen_Timol last updated on 31/May/21

F(x) = ∫_( x) ^( x^2 ) (1/( ln(t) )) dt Show that : ∣ F(x) ∣ ≤ ((∣ x^2 − x ∣)/(∣ ln(x) ∣)) Please

$$\:\:\:\:\:\:{F}\left({x}\right)\:=\:\int_{\:{x}} ^{\:\:{x}^{\mathrm{2}} } \frac{\mathrm{1}}{\:\mathrm{ln}\left({t}\right)\:}\:\mathrm{d}{t} \\ $$$$\:\:\mathrm{Show}\:\mathrm{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:{F}\left({x}\right)\:\mid\:\:\leqslant\:\:\frac{\mid\:{x}^{\mathrm{2}} \:−\:{x}\:\mid}{\mid\:\mathrm{ln}\left({x}\right)\:\mid} \\ $$$$ \\ $$$$\mathrm{Please} \\ $$

Answered by Boucatchou last updated on 31/May/21

∀ x≤t≤x^2 ,x>0, (1/(lnt))≤(1/(lnx)) ⇒ ∣F(x)∣≤∫_x ^( x^2 ) (1/(∣lnt∣))dt≤(1/(∣lnx∣))∫_x ^( x^2 ) dt ⇒ ∣F(x)≤((x^2 −x^x )/(∣lnx∣))≤((∣x^2 −x∣)/(∣lnx∣))

$$\forall\:{x}\leqslant{t}\leqslant{x}^{\mathrm{2}} \:,{x}>\mathrm{0},\:\:\:\frac{\mathrm{1}}{{lnt}}\leqslant\frac{\mathrm{1}}{{lnx}}\:\:\:\:\:\Rightarrow\:\:\:\:\mid{F}\left({x}\right)\mid\leqslant\int_{{x}} ^{\:{x}^{\mathrm{2}} } \frac{\mathrm{1}}{\mid{lnt}\mid}{dt}\leqslant\frac{\mathrm{1}}{\mid{lnx}\mid}\int_{{x}} ^{\:{x}^{\mathrm{2}} } {dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\mid{F}\left({x}\right)\leqslant\frac{{x}^{\mathrm{2}} −{x}^{{x}} }{\mid{lnx}\mid}\leqslant\frac{\mid{x}^{\mathrm{2}} −{x}\mid}{\mid{lnx}\mid} \\ $$

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