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Question Number 78106 by aliesam last updated on 14/Jan/20

Commented by msup trace by abdo last updated on 14/Jan/20
$let f(x)=x^(1−ξ) ∫_x ^(x+1) sin(t^2 )dt changement t^2 =u give f(x)=x^(1−ξ) ∫_x^2 ^((x+1)^2 ) sin(u)(du/(2(√u))) =(x^(1−ξ) /2) ∫_x^2 ^((x+1)^2 ) ((sinu)/(√u))du ∃ c_x ∈]x^2 ,(x+1)^2 [ / 2f(x)=x^(1−ξ) ×(1/(√c_x ))∫_x^2 ^((x+1)^2 ) sinu du =(x^(1−ξ) /(√c_x )){cos(x+1)^2 −cosx^2 } c_x =λx^2 +(1−λ)(x+1)^2 λ ∈]0,1[ ⇒ f(x)=(1/2)(x^(1−ξ) /(√(λx^2 +(1−λ)(x+1)^2 ))){cos(x+1)^2 −co(x^2 )} ∣f(x)∣≤(x^(1−ξ) /(√(λx^2 +(1+λ)(x^2 +2x+1)))) ⇒∣f(x)∣≤(x^(1−ξ) /(x(√(λ+(1+λ)(1+(2/x)+(1/x^2 )))))) ⇒∣f(x)∣≤(1/(x^ξ (√(λ+(1+λ)(1+(2/x)+(1/x^2 )))))) ξ>0 ⇒lim_(x→+∞) (1/x^ξ ) =0 ⇒ lim_(x→+∞) f(x)=0$
$l e t f (x) = x^{1 - ξ} \int_{x}^{x + 1} s i n (t^{2}) d t$ $c h a n g e m e n t t^{2} = u g i v e$ $f (x) = x^{1 - ξ} \int_{x^{2}}^{{(x + 1)}^{2}} s i n (u) \frac{d u}{2 \sqrt{u}}$ $= \frac{x^{1 - ξ}}{2} \int_{x^{2}}^{{(x + 1)}^{2}} \frac{s i n u}{\sqrt{u}} d u$ $\exists c_{x} \in] x^{2}, {(x + 1)}^{2} [/$ $2 f (x) = x^{1 - ξ} \times \frac{1}{\sqrt{c_{x}}} \int_{x^{2}}^{{(x + 1)}^{2}} s i n u d u$ $= \frac{x^{1 - ξ}}{\sqrt{c_{x}}} {c o s {(x + 1)}^{2} - {c o s x}^{2}}$ $c_{x} = λ x^{2} + (1 - λ) {(x + 1)}^{2} λ \in] 0, 1 [$ $\Rightarrow$ $f (x) = \frac{1}{2} \frac{x^{1 - ξ}}{\sqrt{λ x^{2} + (1 - λ) {(x + 1)}^{2}}} {c o s {(x + 1)}^{2} - c o (x^{2})}$ $∣ f (x) ∣⩽ \frac{x^{1 - ξ}}{\sqrt{λ x^{2} + (1 + λ) (x^{2} + 2 x + 1)}}$ $\Rightarrow∣ f (x) ∣⩽ \frac{x^{1 - ξ}}{x \sqrt{λ + (1 + λ) (1 + \frac{2}{x} + \frac{1}{x^{2}}})}$ $\Rightarrow∣ f (x) ∣⩽ \frac{1}{x^{ξ} \sqrt{λ + (1 + λ) (1 + \frac{2}{x} + \frac{1}{x^{2}})}}$ $ξ > 0 \Rightarrow {l i m}_{x \to + \infty} \frac{1}{x^{ξ}} = 0 \Rightarrow$ ${l i m}_{x \to + \infty} f (x) = 0$
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