Question Number 165273 by mnjuly1970 last updated on 28/Jan/22
$$ \\ $$$$\:\:\:\:\:\:{lim}_{\:{n}\rightarrow\:\infty} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\:{n}\:.\:{e}^{\:\mathrm{1}−\:{x}^{\:\mathrm{2}} } }{\:\mathrm{1}\:+\:{n}^{\:\mathrm{2}} \:{x}^{\:\mathrm{2}} }\:{dx}\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−−−− \\ $$
Answered by mindispower last updated on 28/Jan/22
![∫(n/(1+n^2 x^2 ))dx=arctan(nx) lim_(n→∞) [arctan(nx)e^(1−x^2 ) ]_0 ^1 +2∫_0 ^1 xe^(1−x^2 ) arctan(nx)dx =lim_(n→∞) 2∫_0 ^1 arctan(nx).xe^(1−x^2 ) dx 2∫_0 ^1 xtan^(−1) (nx)e^(1−x^2 ) dx≤2∫_0 ^1 (π/2)xe^(1−x^2 ) =(π/2) we Use “Theorem de Conergence Dominee” its The Name in france in English I dont Know How We Call iT lim_(n→∞) 2∫_0 ^1 xtan^(−1) (nx)e^(1−x^2 ) dx=∫_0 ^1 lim_(n→∞) 2tan^(−1) (nx).xe^(1−x^2 ) dx =∫_0 ^1 (π/2)(2xe^(1−x^2 ) )dx.(π/2)[−e^(1−x^2 ) ]_0 ^1 =(π/2)e](https://www.tinkutara.com/question/Q165289.png)
$$\int\frac{{n}}{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{dx}={arctan}\left({nx}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[{arctan}\left({nx}\right){e}^{\mathrm{1}−{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {xe}^{\mathrm{1}−{x}^{\mathrm{2}} } {arctan}\left({nx}\right){dx} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}2}\int_{\mathrm{0}} ^{\mathrm{1}} {arctan}\left({nx}\right).{xe}^{\mathrm{1}−{x}^{\mathrm{2}} } {dx} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {x}\mathrm{tan}^{−\mathrm{1}} \left({nx}\right){e}^{\mathrm{1}−{x}^{\mathrm{2}} } {dx}\leqslant\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\pi}{\mathrm{2}}{xe}^{\mathrm{1}−{x}^{\mathrm{2}} } =\frac{\pi}{\mathrm{2}} \\ $$$${we}\:{Use}\:“{Theorem}\:{de}\:{Conergence}\:{Dominee}'' \\ $$$${its}\:{The}\:{Name}\:{in}\:{france}\:{in}\:{English}\:{I}\:{dont}\:{Know} \\ $$$${How}\:{We}\:{Call}\:{iT} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}2}\int_{\mathrm{0}} ^{\mathrm{1}} {x}\mathrm{tan}^{−\mathrm{1}} \left({nx}\right){e}^{\mathrm{1}−{x}^{\mathrm{2}} } {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\rightarrow\infty} {\mathrm{lim}2tan}^{−\mathrm{1}} \left({nx}\right).{xe}^{\mathrm{1}−{x}^{\mathrm{2}} } {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\pi}{\mathrm{2}}\left(\mathrm{2}{xe}^{\mathrm{1}−{x}^{\mathrm{2}} } \right){dx}.\frac{\pi}{\mathrm{2}}\left[−{e}^{\mathrm{1}−{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{2}}{e} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 28/Jan/22
$$\:\:{bravo}\:,\:{sir}\:{power}\:….{very}\:{nice} \\ $$$$\:{solution} \\ $$$$\:\:\:\:\mathrm{D}{ominating}\:\:{convergence}\:{theorem}… \\ $$
Commented by mindispower last updated on 02/Feb/22
$${Thanx}\:{Sir} \\ $$$${Have}\:\:{a}\:{Nice}\:{Day} \\ $$$$ \\ $$