Question Number 33845 by prof Abdo imad last updated on 26/Apr/18
$${let}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{arctan}\left(\mathrm{1}\:+{n}\right)}{\sqrt{\mathrm{1}+{x}^{{n}} }}\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{I}_{{n}} \:. \\ $$
Commented by prof Abdo imad last updated on 27/Apr/18
![I_n = ∫_R ((arctan(1+n))/(√(1+x^n ))) χ_([0,1]) (x)dx but f_n (x)= ((arctan(1+n))/(√(1+x^n ))) χ_([0,1]) (x)_(n→+∞) →f(x)=(π/2) on[0,1] so ∫_R f_n (x)dx → ∫_0 ^1 f(x)dx =(π/2) .](Q33917.png)
$${I}_{{n}} =\:\int_{{R}} \:\:\frac{{arctan}\left(\mathrm{1}+{n}\right)}{\sqrt{\mathrm{1}+{x}^{{n}} }}\:\chi_{\left[\mathrm{0},\mathrm{1}\right]} \left({x}\right){dx}\:\:{but} \\ $$$${f}_{{n}} \left({x}\right)=\:\frac{{arctan}\left(\mathrm{1}+{n}\right)}{\sqrt{\mathrm{1}+{x}^{{n}} }}\:\chi_{\left[\mathrm{0},\mathrm{1}\right]} \left({x}\right)_{{n}\rightarrow+\infty} \:\rightarrow{f}\left({x}\right)=\frac{\pi}{\mathrm{2}}\:{on}\left[\mathrm{0},\mathrm{1}\right] \\ $$$${so}\:\int_{{R}} {f}_{{n}} \left({x}\right){dx}\:\rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}\:=\frac{\pi}{\mathrm{2}}\:. \\ $$