Question Number 213169 by MrGaster last updated on 31/Oct/24
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{prove} \\ $$$$\:\:\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{n}}{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{e}^{{x}^{\mathrm{2}} } {dx}=\frac{\pi}{\mathrm{2}}. \\ $$
Answered by Berbere last updated on 31/Oct/24
![∫_> ^1 (n/(1+n^2 x^2 ))e^x^2 dx=[tan^(−1) (nx)e^x^2 ]−∫_0 ^1 2xe^x^2 tan^(−1) (nx)dx =(π/2)e−∫_0 ^1 2xe^x^2 tan^(−1) (nx)dx=((πe)/2)−∫_0 ^1 2xe^x^2 [(π/2)−tan^(−1) ((1/(nx)))]dx =((πe)/2)−(π/2)(e−1)+2∫_0 ^1 xe^x^2 tan^(−1) ((1/(nx))) =(π/2)+B ∀x≥0 tan^(−1) (x)<x ⇒∫_0 ^1 xe^x^2 tan^(−1) ((1/(nx)))dx≤∫_0 ^1 (e^x^2 /n)dx≤(e/n) ⇒lim_(n→∞) B=0⇒ lim_(n→∞) ∫_0 ^1 (n/(1+n^2 x^2 ))e^x^2 dx=(π/2)](https://www.tinkutara.com/question/Q213172.png)
$$\int_{>} ^{\mathrm{1}} \frac{{n}}{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{e}^{{x}^{\mathrm{2}} } {dx}=\left[\mathrm{tan}^{−\mathrm{1}} \left({nx}\right){e}^{{x}^{\mathrm{2}} } \right]−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{xe}^{{x}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left({nx}\right){dx} \\ $$$$=\frac{\pi}{\mathrm{2}}{e}−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{xe}^{{x}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left({nx}\right){dx}=\frac{\pi{e}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{2}{xe}^{{x}^{\mathrm{2}} } \left[\frac{\pi}{\mathrm{2}}−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{nx}}\right)\right]{dx} \\ $$$$=\frac{\pi{e}}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\left({e}−\mathrm{1}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {xe}^{{x}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{nx}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}+{B} \\ $$$$\forall{x}\geqslant\mathrm{0}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)<{x} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} {xe}^{{x}^{\mathrm{2}} } \mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{nx}}\right){dx}\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{e}^{{x}^{\mathrm{2}} } }{{n}}{dx}\leqslant\frac{{e}}{{n}} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{B}=\mathrm{0}\Rightarrow \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{n}}{\mathrm{1}+{n}^{\mathrm{2}} {x}^{\mathrm{2}} }{e}^{{x}^{\mathrm{2}} } {dx}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$