let A_n = ∫_0 ^∞ [n e^(−x) ]dx with n integr natural. 1) calculate A_n 2) find lim_(n→+∞) A


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Question Number 43170 by maxmathsup by imad last updated on 07/Sep/18

$l e t A_{n} = \int_{0}^{\infty} [n e^{- x}] d x w i t h n i n t e g r n a t u r a l .$ $1) c a l c u l a t e A_{n}$ $2) f i n d {l i m}_{n \to + \infty} A_{n} .$
Commented by alex041103 last updated on 08/Sep/18

$i s [] t h e w h o l e p a r t f u n c t i o n$ $e . g . [2.67] = 2 ? ? ?$
Commented by maxmathsup by imad last updated on 08/Sep/18

$l o o k s i r a l e x f o r a l l x f r o m R x = [x] + r w i t h 0 ⩽ r < 1 a n d i f x \in Z$ $[x] = x w e h a v e 2, 67 = 2 + 0, 67 a n d 0, 67 \in [0, 1 [\Rightarrow [2, 67] = 2 . a l s o w e h a v e$ $[x] ⩽ x < [x] + 1 .$
Commented by maxmathsup by imad last updated on 09/Sep/18
$1) changement ne^(−x) =t give e^(−x) =(t/n) ⇒−x =ln(t)−ln(n) ⇒−dx=(dt/t) and A_n = −∫_0 ^n [t](−(dt/t)) = ∫_0 ^n (([t])/t)dt = Σ_(k=0) ^(n−1) ∫_k ^(k+1) (k/t)dt =Σ_(k=0) ^(n−1) k {ln(k+1)−ln(k)} = Σ_(k=1) ^n k{ln(k+1)−ln(k)} ⇒A_n = Σ_(k=1) ^n k ln(((k+1)/k)) . 2) we have A_n =Σ_(k=1) ^n k ln(1+(1/k)) but ln(1+x) = x−(x^2 /2) +o(x^3 ) x−(x^2 /2) ≤ln(1+x)≤x ⇒(1/k)−(1/(2k^2 ))≤ln(1+(1/k))≤(1/k) ⇒ k ln(1+(1/k))≥1− (1/(2k)) ⇒ A_n ≥ Σ_(k=1) ^n (1−(1/(2k))) ⇒A_n ≥ n−(1/2) H_n but H_n = ln(n)+γ +o((1/n)) ⇒n−(1/2) H_n =n−(1/2)ln(n)−(γ/2) +o((1/n)) but lim_(n→+∞) n−((ln(n))/2) =lim_(n→+∞) n(1−((ln(n))/(2n)))=+∞ ⇒lim_(n→+∞) A_n =+∞ .$
$1) c h a n g e m e n t {n e}^{- x} = t g i v e e^{- x} = \frac{t}{n} \Rightarrow - x = l n (t) - l n (n) \Rightarrow - d x = \frac{d t}{t}$ $a n d A_{n} = - \int_{0}^{n} [t] (- \frac{d t}{t}) = \int_{0}^{n} \frac{[t]}{t} d t = \sum_{k = 0}^{n - 1} \int_{k}^{k + 1} \frac{k}{t} d t$ $= \sum_{k = 0}^{n - 1} k {l n (k + 1) - l n (k)} = \sum_{k = 1}^{n} k {l n (k + 1) - l n (k)}$ $\Rightarrow A_{n} = \sum_{k = 1}^{n} k l n (\frac{k + 1}{k}) .$ $2) w e h a v e A_{n} = \sum_{k = 1}^{n} k l n (1 + \frac{1}{k}) b u t$ $l n (1 + x) = x - \frac{x^{2}}{2} + o (x^{3}) x - \frac{x^{2}}{2} ⩽ l n (1 + x) ⩽ x \Rightarrow \frac{1}{k} - \frac{1}{2 k^{2}} ⩽ l n (1 + \frac{1}{k}) ⩽ \frac{1}{k} \Rightarrow$ $k l n (1 + \frac{1}{k}) ⩾ 1 - \frac{1}{2 k} \Rightarrow A_{n} ⩾ \sum_{k = 1}^{n} (1 - \frac{1}{2 k}) \Rightarrow A_{n} ⩾ n - \frac{1}{2} H_{n}$ $b u t H_{n} = l n (n) + γ + o (\frac{1}{n}) \Rightarrow n - \frac{1}{2} H_{n} = n - \frac{1}{2} l n (n) - \frac{γ}{2} + o (\frac{1}{n}) b u t$ ${l i m}_{n \to + \infty} n - \frac{l n (n)}{2} = {l i m}_{n \to + \infty} n (1 - \frac{l n (n)}{2 n}) = + \infty \Rightarrow {l i m}_{n \to + \infty} A_{n} = + \infty .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18

	$\int_{0}^{{l o g}_{e} n} [{n e}^{- x}] d x + \int_{{l o g}_{e} n}^{\infty} [{n e}^{- x}] d x$ $n o w {n e}^{- x} = 1$ $e^{- x} = n^{- 1}$ $e^{x} = n$ $x = {l o g}_{e} n w h e n x = {l o g}_{e} n {n e}^{- x} = 1$ $\int_{0}^{{l o g}_{e} n} 1 \times d x + \int_{0}^{\infty} [{n e}^{- x}] d x$ $=∣ x ∣_{0}^{{l o g}_{e} n} + 0$ $= {l o g}_{e} n$
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