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Question Number 33845 by prof Abdo imad last updated on 26/Apr/18

$$let{I}_n={\int }_0^1\frac{arctan(1+n)}{\sqrt{1+x^n}}find{lim}_{n\to +\mathrm{\infty }}{I}_n.$$

Commented by prof Abdo imad last updated on 27/Apr/18

$$I_n={\int }_R\frac{arctan(1+n)}{\sqrt{1+x^n}}{\chi }_{[0,1]}\left(x\right)dxbut$$$$f_n\left(x\right)=\frac{arctan(1+n)}{\sqrt{1+x^n}}{\chi }_{[0,1]}{\left(x\right)}_{n\to +\mathrm{\infty }}\to f\left(x\right)=\frac{\pi }{2}on[0,1]$$$$so{\int }_R{f}_n\left(x\right)dx\to {\int }_0^{1}f\left(x\right)dx=\frac{\pi }{2}.$$

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