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Question Number 208306 by SANOGO last updated on 10/Jun/24

calcul / lim n→+∞ ∫_0 ^(+∞) f_n (x) f_n (x)= arctan((x/n))e^(−x) dx

$${calcul}\:/\:{lim}\:{n}\rightarrow+\infty\:\int_{\mathrm{0}} ^{+\infty} \:{f}_{{n}} \left({x}\right) \\ $$$$\:{f}_{{n}} \left({x}\right)=\:{arctan}\left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx} \\ $$

Commented by SANOGO last updated on 11/Jun/24

thank you

$${thank}\:{you} \\ $$

Answered by Berbere last updated on 10/Jun/24

f_n (x)≤(π/2)e^(−x) x→(π/2)e^(−x) integrable over R+ lim_(n→∞) tan^(−1) ((x/n))e^(−x) =0 ⇒lim_(n→∞) ∫_0 ^∞ f_n (x)=∫_0 ^∞ lim_(n→∞) tan^(−1) ((x/n))e^(−x) dx=∫_0 ^∞ 0dx=0

$${f}_{{n}} \left({x}\right)\leqslant\frac{\pi}{\mathrm{2}}{e}^{−{x}} \:{x}\rightarrow\frac{\pi}{\mathrm{2}}{e}^{−{x}} \:{integrable}\:{over}\:\mathbb{R}+ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right){e}^{−{x}} =\mathrm{0} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\infty} {f}_{{n}} \left({x}\right)=\int_{\mathrm{0}} ^{\infty} \underset{{n}\rightarrow\infty} {\mathrm{lim}tan}^{−\mathrm{1}} \left(\frac{{x}}{{n}}\right){e}^{−{x}} {dx}=\int_{\mathrm{0}} ^{\infty} \mathrm{0}{dx}=\mathrm{0} \\ $$

Answered by mathzup last updated on 10/Jun/24

arctanu)^′ =(1/(1+u^2 ))=Σ_(n=0) ^∞ (−1)^n u^(2n) ⇒arctanu=Σ_(n=0) ^∞ (((−1)^n u^(2n+1) )/(2n+1)) +c (c=0) =u−(u^3 /3)+(u^5 /5)−... ⇒ arctanu≤u ∀u>0 ⇒∣f_n (x)∣≤(x/n) e^(−x) ⇒ ∫_0 ^∞ ∣f_n (x)∣dx≤(1/n)∫_0 ^∞ x e^(−x) dx but ∫_0 ^∞ xe^(−x) dx=Γ(2)=1!=1 ⇒ ∫_0 ^∞ ∣f_n (x)∣dx≤(1/n) →0 (n→+∞) ⇒ lim_(n→+∞) ∫_0 ^∞ f_n (x)dx =0

$$\left.{arctanu}\right)^{'} =\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:{u}^{\mathrm{2}{n}} \\ $$$$\Rightarrow{arctanu}=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} {u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}\:+{c}\:\left({c}=\mathrm{0}\right) \\ $$$$={u}−\frac{{u}^{\mathrm{3}} }{\mathrm{3}}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}−...\:\Rightarrow\:{arctanu}\leqslant{u}\:\:\forall{u}>\mathrm{0} \\ $$$$\Rightarrow\mid{f}_{{n}} \left({x}\right)\mid\leqslant\frac{{x}}{{n}}\:{e}^{−{x}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\mid{f}_{{n}} \left({x}\right)\mid{dx}\leqslant\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \:{x}\:{e}^{−{x}} \:{dx} \\ $$$${but}\:\int_{\mathrm{0}} ^{\infty} \:{xe}^{−{x}} {dx}=\Gamma\left(\mathrm{2}\right)=\mathrm{1}!=\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \mid{f}_{{n}} \left({x}\right)\mid{dx}\leqslant\frac{\mathrm{1}}{{n}}\:\rightarrow\mathrm{0}\:\:\left({n}\rightarrow+\infty\right)\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \int_{\mathrm{0}} ^{\infty} {f}_{{n}} \left({x}\right){dx}\:=\mathrm{0} \\ $$

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