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](https://www.tinkutara.com/question/Q78132.png)
$${let}\:{f}\left({x}\right)={x}^{\mathrm{1}−\xi} \:\int_{{x}} ^{{x}+\mathrm{1}} {sin}\left({t}^{\mathrm{2}} \right){dt} \\ $$$${changement}\:{t}^{\mathrm{2}} ={u}\:{give} \\ $$$${f}\left({x}\right)={x}^{\mathrm{1}−\xi} \:\:\:\int_{{x}^{\mathrm{2}} } ^{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \:{sin}\left({u}\right)\frac{{du}}{\mathrm{2}\sqrt{{u}}} \\ $$$$=\frac{{x}^{\mathrm{1}−\xi} }{\mathrm{2}}\:\int_{{x}^{\mathrm{2}} } ^{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \:\:\frac{{sinu}}{\:\sqrt{{u}}}{du} \\ $$$$\left.\exists\:{c}_{{x}} \:\in\right]{x}^{\mathrm{2}} ,\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left[\:\:/\right. \\ $$$$\mathrm{2}{f}\left({x}\right)={x}^{\mathrm{1}−\xi} \:×\frac{\mathrm{1}}{\:\sqrt{{c}_{{x}} }}\int_{{x}^{\mathrm{2}} } ^{\left({x}+\mathrm{1}\right)^{\mathrm{2}} } {sinu}\:{du} \\ $$$$=\frac{{x}^{\mathrm{1}−\xi} }{\:\sqrt{{c}_{{x}} }}\left\{{cos}\left({x}+\mathrm{1}\right)^{\mathrm{2}} −{cosx}^{\mathrm{2}} \right\} \\ $$$$\left.{c}_{{x}} =\lambda{x}^{\mathrm{2}} \:+\left(\mathrm{1}−\lambda\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:\:\lambda\:\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$$$\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\frac{{x}^{\mathrm{1}−\xi} }{\:\sqrt{\lambda{x}^{\mathrm{2}} \:+\left(\mathrm{1}−\lambda\right)\left({x}+\mathrm{1}\right)^{\mathrm{2}} }}\left\{{cos}\left({x}+\mathrm{1}\right)^{\mathrm{2}} −{co}\left({x}^{\mathrm{2}} \right)\right\} \\ $$$$\mid{f}\left({x}\right)\mid\leqslant\frac{{x}^{\mathrm{1}−\xi} }{\:\sqrt{\lambda{x}^{\mathrm{2}} +\left(\mathrm{1}+\lambda\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)}} \\ $$$$\Rightarrow\mid{f}\left({x}\right)\mid\leqslant\frac{{x}^{\mathrm{1}−\xi} }{\left.{x}\sqrt{\lambda+\left(\mathrm{1}+\lambda\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right.}\right)} \\ $$$$\Rightarrow\mid{f}\left({x}\right)\mid\leqslant\frac{\mathrm{1}}{{x}^{\xi} \sqrt{\lambda+\left(\mathrm{1}+\lambda\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}} \\ $$$$\xi>\mathrm{0}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} \:\:\frac{\mathrm{1}}{{x}^{\xi} }\:=\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} \:{f}\left({x}\right)=\mathrm{0} \\ $$