find lim ∫_(1/n) ^n arctan(1+(x/n))e^(−nx) dx


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Question Number 121315 by Bird last updated on 06/Nov/20

$f i n d l i m \int_{\frac{1}{n}}^{n} a r c t a n (1 + \frac{x}{n}) e^{- n x} d x$
	Answered by Lordose last updated on 06/Nov/20

	$u = (1 + \frac{x}{n}) \Rightarrow du = \frac{dx}{n}$ $x = n (u - 1)$ $\int_{\frac{n^{2} + 1}{n^{2}}}^{2} \tan^{- 1} (u) e^{- n^{2} (u - 1)} du$ $\int_{\frac{n^{2} + 1}{n^{2}}}^{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} u^{2 n + 1}}{(2 n + 1)} e^{- n^{2} (u - 1)} du$ $e^{n^{2}} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{(2 n + 1)} \int_{\frac{n^{2} + 1}{n^{2}}}^{2} u^{2 n + 1} e^{- n^{2} u} du$ $Wait$
	Answered by mathmax by abdo last updated on 06/Nov/20
	$A_n =∫_(1/n) ^n arctan(1+(x/n))e^(−nx) dx changement (x/n)=t give A_n =∫_(1/n^2 ) ^1 arctan(1+t)e^(−n^2 t) ndt =n ∫_(1/n^2 ) ^1 arctan(1+t)e^(−n^2 t) dt ⇒(A_n /n) =_(by parts) [−(1/n^2 )e^(−n^2 t) arctan(1+t)]_(1/n^2 ) ^1 +(1/n^2 )∫_(1/n^2 ) ^1 (e^(−n^2 t) /(1+(1+t)^2 ))dt =(1/n^2 )e^(−1) arctan(1+(1/n^2 ))−(1/n^2 )e^(−n^2 ) arctan(2)+(1/n^2 )∫_(1/n^2 ) ^1 (e^(−n^2 t) /(1+(1+t)^2 ))dt ⇒ A_n =(1/n)e^(−1) arctan(1+(1/n^2 ))−(1/n)e^(−n^2 ) arctan(2)+(1/n)∫_(1/n^2 ) ^1 (e^(−n^2 t) /(1+(1+t)^2 ))dt we have lim_(n→+∞) (1/n){e^(−1) arctan(1+(1/n^2 ))−e^(−n^2 arctan2) }=0 ∫_(1/n^2 ) ^1 (e^(−n^2 t) /(1+(1+t)^2 ))dt =∫_R f_n (x)dx with f_n (x)=(e^(−n^2 t) /(1+(1+t)^2 )) χ_(](1/n^2 ),1]) (t)dt f_n →^(cs) 0 ⇒(1/n)∫_R f_n (x)dx→0 ⇒lim_(n→+∞) A_n =0$
	$A_{n} = \int_{\frac{1}{n}}^{n} \arctan (1 + \frac{x}{n}) e^{- nx} dx changement \frac{x}{n} = t give$ $A_{n} = \int_{\frac{1}{n^{2}}}^{1} \arctan (1 + t) e^{- n^{2} t} ndt = n \int_{\frac{1}{n^{2}}}^{1} \arctan (1 + t) e^{- n^{2} t} dt$ $\Rightarrow \frac{A_{n}}{n} =_{by parts} {[- \frac{1}{n^{2}} e^{- n^{2} t} \arctan (1 + t)]}_{\frac{1}{n^{2}}}^{1} + \frac{1}{n^{2}} \int_{\frac{1}{n^{2}}}^{1} \frac{e^{- n^{2} t}}{1 + {(1 + t)}^{2}} dt$ $= \frac{1}{n^{2}} e^{- 1} \arctan (1 + \frac{1}{n^{2}}) - \frac{1}{n^{2}} e^{- n^{2}} \arctan (2) + \frac{1}{n^{2}} \int_{\frac{1}{n^{2}}}^{1} \frac{e^{- n^{2} t}}{1 + {(1 + t)}^{2}} dt \Rightarrow$ $A_{n} = \frac{1}{n} e^{- 1} \arctan (1 + \frac{1}{n^{2}}) - \frac{1}{n} e^{- n^{2}} \arctan (2) + \frac{1}{n} \int_{\frac{1}{n^{2}}}^{1} \frac{e^{- n^{2} t}}{1 + {(1 + t)}^{2}} dt$ $we have \lim_{n \to + \infty} \frac{1}{n} {e^{- 1} \arctan (1 + \frac{1}{n^{2}}) - e^{- n^{2} \arctan 2}} = 0$ $\int_{\frac{1}{n^{2}}}^{1} \frac{e^{- n^{2} t}}{1 + {(1 + t)}^{2}} dt = \int_{R} f_{n} (x) dx with f_{n} (x) = \frac{e^{- n^{2} t}}{1 + {(1 + t)}^{2}} χ_{] \frac{1}{n^{2}}, 1]} (t) dt$ $f_{n} \to^{cs} 0 \Rightarrow \frac{1}{n} \int_{R} f_{n} (x) dx \to 0 \Rightarrow \lim_{n \to + \infty} A_{n} = 0$
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