Question Number 171684 by SANOGO last updated on 19/Jun/22
Answered by mindispower last updated on 19/Jun/22
![f(−x)=f(x)⇔etudier f sur R_+ ∀x∈R_+ ∣F(x)∣≤∫_1 ^∞ ((ln(t))/(x^2 +t^2 ))dt≤∫_1 ^∞ ((ln(t))/t^2 )dt ∫_1 ^∞ ((ln(t))/t^2 )dt<∞ utilsons ln(t)=o((√t)),t→∞ ((√t)/t^2 )=(1/t^(3/2) ) integrable en +∞⇒∫_1 ^∞ ((ln(t))/t^2 )dt exist ⇒F(x) est definit sur R 2 evident ln(t)≥0,∀t∈[1,∞] F(0)=∫_1 ^∞ ((ln(t))/t^2 )dt=[−((ln(t))/t)]_1 ^∞ +∫(1/t^2 )dt =[−(1/t)]_1 ^∞ =1 3,∀x∈R_+ t→((ln(t))/(x^2 +t^2 )) est continue sur [1,∞[ ∫_0 ^1 ((ln(t))/(x^2 +t^2 ))dt≤∫_0 ^1 ((ln(t))/t^2 )dt=F(0)=1 ⇒∀x∈R ∫_1 ^∞ ((ln(t))/(x^2 +t^2 ))dt est continue](Q171687.png)
$${f}\left(−{x}\right)={f}\left({x}\right)\Leftrightarrow{etudier}\:{f}\:{sur}\:{R}_{+} \\ $$$$\forall{x}\in{R}_{+} \\ $$$$\mid{F}\left({x}\right)\mid\leqslant\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({t}\right)}{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}\leqslant\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} }{dt} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} }{dt}<\infty\:{utilsons}\:\:\:{ln}\left({t}\right)={o}\left(\sqrt{{t}}\right),{t}\rightarrow\infty \\ $$$$\frac{\sqrt{{t}}}{{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{integrable}\:{en}\:+\infty\Rightarrow\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} }{dt}\:{exist} \\ $$$$\Rightarrow{F}\left({x}\right)\:{est}\:{definit}\:{sur}\:\mathbb{R} \\ $$$$\mathrm{2}\:{evident}\:{ln}\left({t}\right)\geqslant\mathrm{0},\forall{t}\in\left[\mathrm{1},\infty\right] \\ $$$${F}\left(\mathrm{0}\right)=\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} }{dt}=\left[−\frac{{ln}\left({t}\right)}{{t}}\right]_{\mathrm{1}} ^{\infty} +\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$=\left[−\frac{\mathrm{1}}{{t}}\right]_{\mathrm{1}} ^{\infty} =\mathrm{1} \\ $$$$\mathrm{3},\forall{x}\in\mathbb{R}_{+} {t}\rightarrow\frac{{ln}\left({t}\right)}{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} }\:{est}\:{continue}\:{sur}\:\left[\mathrm{1},\infty\left[\right.\right. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{{t}^{\mathrm{2}} }{dt}={F}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\Rightarrow\forall{x}\in\mathbb{R}\:\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({t}\right)}{{x}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}\:{est}\:{continue} \\ $$
Commented by SANOGO last updated on 20/Jun/22
$${bien}\:{merci} \\ $$
Commented by mindispower last updated on 20/Jun/22
$${avec}\:{plaisir} \\ $$