find lim_(n→+∞) ∫_0 ^n e^(nx) arctan((x/n))dx


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Question Number 206730 by mathzup last updated on 23/Apr/24

$f i n d {l i m}_{n \to + \infty} \int_{0}^{n} e^{n x} a r c t a n (\frac{x}{n}) d x$
Commented by aleks041103 last updated on 24/Apr/24

$M o r e i n t e r e s t i n g i s$ $\underset{n \to \infty}{l i m} \int_{0}^{n} e^{- n x} a r c t a n (\frac{x}{n}) d x = \underset{n \to \infty}{l i m} J_{n}$ $J_{n} = \int_{0}^{n} e^{- n x} a r c t a n (\frac{x}{n}) d x =$ $= \int_{0}^{1} {n e}^{- n^{2} t} a r c t a n (t) d t = \int_{0}^{1} f_{n} (t) d t$ $o b v i o u s l y i f w e f i x t t h e n :$ $\underset{n \to \infty}{l i m} f_{n} (t) = a r c t a n (t) \underset{n \to \infty}{l i m} {n e}^{- n^{2} t} = 0$ $\underset{n \to \infty}{l i m} f_{n} (t) = 0, t \in [0, 1]$ $\Rightarrow f_{n} \overset{[0, 1]}{\to} 0 p o i n t w i s e$ $b u t a l s o :$ $a r c t a n (t) ⩽ t f o r t \in [0, 1]$ $\Rightarrow 0 ⩽ f_{n} (t) ⩽ {n t e}^{- n^{2} t} = \frac{1}{n} ({a e}^{- a}), a = n^{2} t \in [0, n^{2}] \subset [0, \infty)$ $\Rightarrow f_{n} (t) ⩽ \frac{1}{n} ({a e}^{- a}) ⩽ \frac{1}{n} (\underset{a \in [0, n^{2}]}{s u p} {a e}^{- a}) ⩽ \frac{1}{n} (\underset{a ⩾ 0}{s u p} {a e}^{- a})$ $b u t$ $g (a) = {a e}^{- a} \Rightarrow g^{'} = (1 - a) e^{- a} = 0 \Rightarrow a = 1$ $\Rightarrow \underset{a ⩾ 0}{s u p} {a e}^{- a} = g (1) = e^{- 1}$ $\Rightarrow 0 ⩽ f_{n} (t) ⩽ \frac{1}{e n}, t \in [0, 1]$ $\Rightarrow∥ f_{n} ∥= \underset{t \in [0, 1]}{s u p} ∣ f_{n} (t) ∣⩽ \frac{1}{e n}$ $\Rightarrow \underset{n \to \infty}{l i m} ∥ f_{n} ∥= 0$ $\Rightarrow f_{n} \overset{[0, 1]}{⇉} 0, i . e . u n i f o r m l y$ $\Rightarrow \underset{n \to \infty}{l i m} \int_{0}^{1} f_{n} (t) d t = \int_{0}^{1} [\underset{n \to \infty}{l i m} f_{n} (t)] d t =$ $= \int_{0}^{1} 0 d t = 0$ $\Rightarrow \underset{n \to \infty}{l i m} J_{n} = 0$ $\Rightarrow \underset{n \to \infty}{l i m} \int_{0}^{n} e^{- n x} a r c t a n (\frac{x}{n}) d x = 0$
	Answered by aleks041103 last updated on 24/Apr/24
	$x=nt⇒dx=ndt ⇒∫_0 ^( n) e^(nx) arctan(x/n)dx=∫_0 ^( 1) e^(n^2 t) arctan(t)ndt=I_n but e^(n^2 t) ≥1 for t∈[0,1] I_n =n∫_0 ^( 1) e^(n^2 t) arctan(t)dt≥n∫_0 ^( 1) arctan(t)dt= =n{[t arctan(t)]_0 ^1 −∫_0 ^( 1) ((tdt)/(1+t^2 ))}= =n{(π/4)−(1/2)[ln(1+t^2 )]_0 ^1 }=n((π/4)−((ln(2))/2)) but (π/4)>(3/4)=0.75 and (1/2)ln(2)<(1/2)=0.5 ⇒c=(π/4)−((ln(2))/2)>0.25>0 ⇒I_n >cn, c>0 ⇒since lim_(n→∞) cn = +∞ then lim_(n→∞) I_n = lim_(n→∞) ∫_0 ^( n) e^(nx) arctan((x/n))dx = +∞$
	$x = n t \Rightarrow d x = n d t$ $\Rightarrow \int_{0}^{n} e^{n x} a r c t a n (x / n) d x = \int_{0}^{1} e^{n^{2} t} a r c t a n (t) n d t = I_{n}$ $b u t e^{n^{2} t} ⩾ 1 f o r t \in [0, 1]$ $I_{n} = n \int_{0}^{1} e^{n^{2} t} a r c t a n (t) d t ⩾ n \int_{0}^{1} a r c t a n (t) d t =$ $= n {{[t a r c t a n (t)]}_{0}^{1} - \int_{0}^{1} \frac{t d t}{1 + t^{2}}} =$ $= n {\frac{π}{4} - \frac{1}{2} {[l n (1 + t^{2})]}_{0}^{1}} = n (\frac{π}{4} - \frac{l n (2)}{2})$ $b u t \frac{π}{4} > \frac{3}{4} = 0.75 a n d \frac{1}{2} l n (2) < \frac{1}{2} = 0.5$ $\Rightarrow c = \frac{π}{4} - \frac{l n (2)}{2} > 0.25 > 0$ $\Rightarrow I_{n} > c n, c > 0$ $\Rightarrow s i n c e \underset{n \to \infty}{l i m} c n = + \infty t h e n$ $\underset{n \to \infty}{l i m} I_{n} = \underset{n \to \infty}{l i m} \int_{0}^{n} e^{n x} a r c t a n (\frac{x}{n}) d x = + \infty$
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