let-A-n-0-n-e-x-dx-with-n-integr-natural-1-calculate-A-n-2-find-lim-n-A-n-

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) 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