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Question Number 91715 by 174 last updated on 02/May/20

Commented by mathmax by abdo last updated on 02/May/20

$$let{A}_n={\int }_0^{\frac{1}{\sqrt{n}}}\frac{n\sqrt{1+x^4}}{1+n^2{x}^2}dxchangementnx=tgive$$$$A_n={\int }_0^{\sqrt{n}}\frac{n\sqrt{1+\frac{t^4}{n^4}}}{1+t^2}\frac{dt}{n}={\int }_0^{\sqrt{n}}\frac{\sqrt{1+\frac{t^4}{n^4}}}{1+t^2}dt$$$$={\int }_R{\phi }_n\left(t\right)dtwith{\phi }_n\left(t\right)=\frac{\sqrt{1+\frac{t^4}{n^4}}}{1+t^2}{\chi }_{[0,\sqrt{n}}[\left(t\right)$$$${\phi }_n{\to }^{cs}\phi =\frac{1}{1+t^2}and\mid {\phi }_n\left(t\right)\mid ⩽\frac{1}{1+t^2}\mathrm{\forall }t\in [0,\sqrt{n}[$$$$theoremofconv.domineegive$$$${lim}_{n\to +\mathrm{\infty }}{A}_n={lim}_{n\to +\mathrm{\infty }}{\int }_R{\phi }_n\left(t\right)dt={\int }_R{lim}_{n\to \mathrm{\infty }}{\phi }_n\left(t\right)dt$$$$={\int }_0^{\mathrm{\infty }}\frac{dt}{1+t^2}=\frac{\pi }{2}$$

Commented by 174 last updated on 02/May/20

thanks

Commented by abdomathmax last updated on 02/May/20

$$youarewelcome$$

Commented by abdomathmax last updated on 02/May/20

$$erroroftyppo\mid {\phi }_n\mid ⩽\frac{\sqrt{2}}{1+t^2}$$

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