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let-give-f-x-x-2-cos-1-x-2-if-x-0-1-but-its-derivative-f-is-not-integrable-on-0-1-

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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].
letgivef(x)=x2cos(1x2)ifx∈]0,1]butitsderivativef′isnotintegrableon]0,1].
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 .
wehavelimx→0f(x)=0because∣x2cos(1x)∣⩽x2limx→0f(x)x=limx→0xcos(1x)=0sofisderivableon[0,1]fromanothersidef′(x)=2xcos(1x2)−x2(−2x−3)sin(1x2)=2xcos(1x2)−2xsin(1x2))and0∫01f′(x)dx=2∫01xcos(1x2)dx−2∫011xsin(1x2)ch1x=t∫01xcos(1x2)dx=−∫1+∞1tcos(t2)(−dtt2)=∫1+∞cos(t2)t3dtintegralconv.and∫011xsin(1x2)=−∫1+∞tsint2−dtt2=∫1+∞sin(t2)tdtandthisintegralis?divergent.

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