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Question Number 29554 by abdo imad last updated on 09/Feb/18
![let give f(x)= x^2 cos((1/x^2 )) if x∈]0,1] but its derivative f^′ is not integrable on ]0,1].](Q29554.png)
$$\left.{l}\left.{et}\:{give}\:{f}\left({x}\right)=\:{x}^{\mathrm{2}} {cos}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:{if}\:{x}\in\right]\mathrm{0},\mathrm{1}\right]\:{but}\:{its}\:{derivative}\:{f}^{'} \\ $$$$\left.{i}\left.{s}\:{not}\:{integrable}\:{on}\:\right]\mathrm{0},\mathrm{1}\right]. \\ $$
Commented by abdo imad last updated on 14/Feb/18
![we have lim_(x→0) f(x)=0 because ∣x^2 cos((1/x))∣≤ x^2 lim_(x→0) ((f(x))/x) =lim_(x→0) xcos((1/x))=0 so f is derivable on [0,1] from another side f^′ (x)= 2xcos((1/x^2 ))−x^2 (−2x^(−3) )sin((1/x^2 )) =2xcos((1/x^2 )) −(2/x) sin((1/x^(2)) ))and_0_ ∫_0 ^1 f^′ (x)dx = 2 ∫_0 ^1 xcos((1/x^2 ))dx −2∫_0 ^1 (1/x)sin((1/x^2 )) ch (1/x)=t ∫_0 ^1 xcos((1/x^2 ))dx =−∫_1 ^(+∞) (1/t)cos(t^2 )(−(dt/t^2 )) =∫_1 ^(+∞) ((cos(t^2 ))/t^3 )dt integral conv.and ∫_0 ^1 (1/x) sin((1/x^2 ))=−∫_1 ^(+∞) t sint^2 ((−dt)/t^2 )= ∫_1 ^(+∞) ((sin(t^2 ))/t)dt and this integral is?divergent .](Q29947.png)
$${we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}} {f}\left({x}\right)=\mathrm{0}\:{because}\:\mid{x}^{\mathrm{2}} {cos}\left(\frac{\mathrm{1}}{{x}}\right)\mid\leqslant\:{x}^{\mathrm{2}} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{f}\left({x}\right)}{{x}}\:={lim}_{{x}\rightarrow\mathrm{0}} \:{xcos}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{0}\:{so}\:{f}\:{is}\:{derivable}\:{on} \\ $$$$\left[\mathrm{0},\mathrm{1}\right]\:\:{from}\:{another}\:{side}\:{f}^{'} \left({x}\right)=\:\mathrm{2}{xcos}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)−{x}^{\mathrm{2}} \left(−\mathrm{2}{x}^{−\mathrm{3}} \right){sin}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right) \\ $$$$=\mathrm{2}{xcos}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:−\frac{\mathrm{2}}{{x}}\:{sin}\left(\frac{\mathrm{1}}{{x}^{\left.\mathrm{2}\right)} }\right){and}_{\mathrm{0}_{} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{f}^{'} \left({x}\right){dx}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{xcos}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}\:−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{x}}{sin}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\:{ch}\:\frac{\mathrm{1}}{{x}}={t} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{xcos}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}\:=−\int_{\mathrm{1}} ^{+\infty} \frac{\mathrm{1}}{{t}}{cos}\left({t}^{\mathrm{2}} \right)\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right) \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:\frac{{cos}\left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{3}} }{dt}\:{integral}\:{conv}.{and} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{x}}\:{sin}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=−\int_{\mathrm{1}} ^{+\infty} \:{t}\:{sint}^{\mathrm{2}} \:\frac{−{dt}}{{t}^{\mathrm{2}} }=\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{{t}}{dt}\:{and} \\ $$$${this}\:{integral}\:{is}?{divergent}\:. \\ $$