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Question Number 80882 by mr W last updated on 07/Feb/20

Commented by jagoll last updated on 07/Feb/20

$i n c o m p l e x a n a l y s i s ?$
Commented by ajfour last updated on 07/Feb/20

$= 1 ?$
Commented by mr W last updated on 07/Feb/20

$y e s, i t h i n k s o .$ $\lim_{n \to \infty} \int_{0}^{\sqrt{3}} \frac{1}{1 + x^{n}} d x = \lim_{n \to \infty} \int_{0}^{15} \frac{1}{1 + x^{n}} d x$ $= \lim_{n \to \infty} \int_{0}^{10000000000000} \frac{1}{1 + x^{n}} d x = 1$ $e v e n \lim_{n \to \infty} \int_{0}^{\infty} \frac{1}{1 + x^{n}} d x = 1 ?$
Commented by mathmax by abdo last updated on 07/Feb/20

$l e t I_{n} = \int_{0}^{\sqrt{3}} \frac{d x}{1 + x^{n}} \Rightarrow I_{n} = \int_{R} \frac{1}{1 + x^{n}} χ_{[0, \sqrt{3]}} (x) d x = \int_{R} f_{n} (x) d x$ $w i t h f_{n} (x) = \frac{1}{1 + x^{n}} χ_{[0, \sqrt{3}]} (x) d x$ $i f 0 ⩽ x < 1 f_{n} \to χ_{[0, \sqrt{3}]} (x) = 1 \Rightarrow {l i m}_{n \to + \infty} I_{n} = \sqrt{3}$ $i f x > 1 f_{n} \to 0 (s . c) a n d w e h a v e x^{n} > 1 \Rightarrow \frac{1}{1 + x^{n}} χ_{[0, \sqrt{3}]} (x) < \frac{1}{2} χ_{[0, \sqrt{3}]} (x)$ $t h e o r e m o f c o n v e r g e n c e d o m i n e e g i v e$ ${l i m}_{n \to + \infty} \int_{R} f_{n} (x) d x = \int_{R} {l i m}_{n \to + \infty} f_{n} (x) d x = 0$
	Answered by ~blr237~ last updated on 07/Feb/20
	$let g(a)= lim_(n→∞) ∫_0 ^a f_n (x) dx with f_n (x)=(1/(1+x^n )) ,a>0 we have ∀ x∈[0,a] 0<(a/(1+a^n ))≤∫_0 ^a (1/(1+x^n ))dx ≤ a so ∀ n f_n is integrable and (∫_0 ^a f_n (x)dx)_(n≥1) is confined if x∈[0,1] x^(n+1) ≤x^n and f_n (x)≤f_(n+1) (x) if x∈[1:∞[ x^n <x^(n+1) and f_(n+1) (x)≤f_n (x) then (f_n ) is monotone :so due to CSM lim_(n→∞) ∫_0 ^a f_n (x)dx=∫_0 ^a lim_(n→∞) f_n (x)dx= { ((∫_0 ^a 1dx if a<1)),((∫_0 ^1 1dx +∫_1 ^a 0dx if a≥1)) :}= { ((a if a<1)),((1 if a≥1)) :} lim_(n→∞) ∫_0 ^∞ (1/(1+x^n ))dx=lim_(a→∞) g(a) = 1$
	$l e t g (a) = \lim_{n \to \infty} \int_{0}^{a} f_{n} (x) d x w i t h f_{n} (x) = \frac{1}{1 + x^{n}}, a > 0$ $w e h a v e \forall x \in [0, a] 0 < \frac{a}{1 + a^{n}} ⩽ \int_{0}^{a} \frac{1}{1 + x^{n}} d x ⩽ a$ $s o \forall n f_{n} i s i n t e g r a b l e a n d {(\int_{0}^{a} f_{n} (x) d x)}_{n ⩾ 1} i s c o n f i n e d$ $i f x \in [0, 1] x^{n + 1} ⩽ x^{n} a n d f_{n} (x) ⩽ f_{n + 1} (x)$ $i f x \in [1 : \infty [x^{n} < x^{n + 1} a n d f_{n + 1} (x) ⩽ f_{n} (x)$ $t h e n (f_{n}) i s m o n o t o n e : s o d u e t o C S M$ $\lim_{n \to \infty} \int_{0}^{a} f_{n} (x) d x = \int_{0}^{a} \lim_{n \to \infty} f_{n} (x) d x = {\begin{cases} \int_{0}^{a} 1 d x i f a < 1 \\ \int_{0}^{1} 1 d x + \int_{1}^{a} 0 d x i f a ⩾ 1 \end{cases} = {\begin{cases} a i f a < 1 \\ 1 i f a ⩾ 1 \end{cases}$ $\lim_{n \to \infty} \int_{0}^{\infty} \frac{1}{1 + x^{n}} d x = \lim_{a \to \infty} g (a) = 1$
	Commented by mr W last updated on 07/Feb/20

	$t h a n k s a l o t s i r!$
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