find-lim-n-0-n-e-nx-arctan-x-n-dx-

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 = 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