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

$$\underset{n\to \mathrm{\infty }}{\lim }n{\int }_0^{\mathrm{\infty }}\sin x^n\mathrm{d}x$$

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

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

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

$$let{I}_n=n{\int }_0^{\mathrm{\infty }}sin\left(x^n\right)dx{x}^n=t\Rightarrow x=t^{\frac{1}{n}}\Rightarrow {\int }_0^{\mathrm{\infty }}sin\left(x^n\right)dx$$$$={\int }_0^{\mathrm{\infty }}sin\left(t\right)\frac{1}{n}{t}^{\frac{1}{n}-1}dt=\frac{1}{n}{\int }_0^{\mathrm{\infty }}{t}^{\frac{1}{n}-1}sintdt\Rightarrow$$$$I_n={\int }_0^{\mathrm{\infty }}{t}^{\frac{1}{n}-1}sin\left(t\right)dt={\int }_R{t}^{\frac{1}{n}-1}sint{\chi }_{[0,+\mathrm{\infty }[}\left(t\right)dt$$$${\to }_{n\to +\mathrm{\infty }}{\int }_0^{+\mathrm{\infty }}\frac{sin\left(t\right)}{t}dt=\frac{\pi }{2}.$$

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