Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 143622 by ArielVyny last updated on 16/Jun/21

∫_0 ^∝ e^(2arctg(t^2 )) dt

$$\int_{\mathrm{0}} ^{\propto} {e}^{\mathrm{2}{arctg}\left({t}^{\mathrm{2}} \right)} {dt} \\ $$

Answered by TheHoneyCat last updated on 17/Jun/21

arctan(t)→_(t→+∞) (π/2)>0 so e^(2arctan(t^2 )) →_(t→+∞) e^π >e^3 >e>1 thus ∃X∈R_+ ∣ ∀x∈[x,+∞[ e^(2arctan(x^2 )) >1 let X be such a real number ∫_0 ^(+∞) e^(2arctan(t^2 )) dt=∫_0 ^X e^(2arctan(t^2 )) dt + ∫_X ^(+∞) e^(2arctan(t^2 )) dt ∫_0 ^X e^(2arctan(t^2 )) dt ∈R as the integral of a Continued function on a segment ∫_X ^(+∞) e^(2arctan(t^2 )) dt ≥ ∫_X ^(+∞) dt=lim_(x→+∞) (x−X)=+∞ in a word: ∫_0 ^(+∞) e^(2artcan(t^2 )) dt=+∞ _■ perhaps you made a mistake when writting it...

$$\mathrm{arctan}\left({t}\right)\underset{{t}\rightarrow+\infty} {\rightarrow}\frac{\pi}{\mathrm{2}}>\mathrm{0} \\ $$$$\mathrm{so}\:{e}^{\mathrm{2arctan}\left({t}^{\mathrm{2}} \right)} \underset{{t}\rightarrow+\infty} {\rightarrow}{e}^{\pi} >{e}^{\mathrm{3}} >{e}>\mathrm{1} \\ $$$$\mathrm{thus}\:\exists\mathrm{X}\in\mathbb{R}_{+} \mid\:\forall{x}\in\left[{x},+\infty\left[\:{e}^{\mathrm{2arctan}\left({x}^{\mathrm{2}} \right)} >\mathrm{1}\right.\right. \\ $$$$\mathrm{let}\:\mathrm{X}\:\:\mathrm{be}\:\mathrm{such}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number} \\ $$$$\int_{\mathrm{0}} ^{+\infty} {e}^{\mathrm{2arctan}\left({t}^{\mathrm{2}} \right)} {dt}=\int_{\mathrm{0}} ^{\mathrm{X}} {e}^{\mathrm{2arctan}\left({t}^{\mathrm{2}} \right)} {dt}\:+\:\int_{\mathrm{X}} ^{+\infty} {e}^{\mathrm{2arctan}\left({t}^{\mathrm{2}} \right)} {dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{X}} {e}^{\mathrm{2arctan}\left({t}^{\mathrm{2}} \right)} {dt}\:\in\mathbb{R}\:\mathrm{as}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{of}\:\mathrm{a}\:\mathrm{Continued}\:\mathrm{function}\:\mathrm{on}\:\mathrm{a}\:\mathrm{segment} \\ $$$$\int_{\mathrm{X}} ^{+\infty} {e}^{\mathrm{2arctan}\left({t}^{\mathrm{2}} \right)} {dt}\:\geqslant\:\int_{\mathrm{X}} ^{+\infty} {dt}=\mathrm{lim}_{{x}\rightarrow+\infty} \left({x}−\mathrm{X}\right)=+\infty \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{word}:\:\int_{\mathrm{0}} ^{+\infty} {e}^{\mathrm{2artcan}\left({t}^{\mathrm{2}} \right)} {dt}=+\infty\:_{\blacksquare} \\ $$$$ \\ $$$${perhaps}\:{you}\:{made}\:{a}\:{mistake}\:{when}\:{writting}\:{it}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com