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Question Number 19920 by lidaye last updated on 18/Aug/17

lim_(n→∞) n∫_0 ^∞ sin x^n dx

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{n}\int_{\mathrm{0}} ^{\infty} \mathrm{sin}\:{x}^{{n}} \mathrm{d}{x} \\ $$

Commented by 1kanika# last updated on 18/Aug/17

what is the answer of this question?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{of}\:\mathrm{this}\:\mathrm{question}? \\ $$

Commented by prof Abdo imad last updated on 22/Jun/18

let I_n =n∫_0 ^∞ sin(x^n )dx x^n =t ⇒x=t^(1/n) ⇒∫_0 ^∞ sin(x^n )dx =∫_0 ^∞ sin(t) (1/n)t^((1/n) −1) dt =(1/n) ∫_0 ^∞ t^((1/n) −1) sint dt ⇒ I_n = ∫_0 ^∞ t^((1/n)−1) sin(t)dt= ∫_R t^((1/n)−1) sint χ_([0,+∞[) (t)dt →_(n→+∞) ∫_0 ^(+∞) ((sin(t))/t) dt =(π/2) .

$${let}\:{I}_{{n}} ={n}\int_{\mathrm{0}} ^{\infty} {sin}\left({x}^{{n}} \right){dx}\:{x}^{{n}} \:={t}\:\Rightarrow{x}={t}^{\frac{\mathrm{1}}{{n}}} \:\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:{sin}\left({x}^{{n}} \right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:{sin}\left({t}\right)\:\frac{\mathrm{1}}{{n}}{t}^{\frac{\mathrm{1}}{{n}}\:−\mathrm{1}} {dt}\:=\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\infty} \:\:{t}^{\frac{\mathrm{1}}{{n}}\:−\mathrm{1}} \:{sint}\:{dt}\:\Rightarrow \\ $$$${I}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:{sin}\left({t}\right){dt}=\:\int_{{R}} \:{t}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:{sint}\:\chi_{\left[\mathrm{0},+\infty\left[\right.\right.} \left({t}\right){dt} \\ $$$$\rightarrow_{{n}\rightarrow+\infty} \:\:\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{sin}\left({t}\right)}{{t}}\:{dt}\:=\frac{\pi}{\mathrm{2}}\:. \\ $$

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