calculatelim-n-0-1-x-n-n-ln-1-2x-dx-

Question Number 100514 by mathmax by abdo last updated on 27/Jun/20

{calculatelim}_{n \to + \infty} \int_{0}^{\infty} {(1 - \frac{x}{n})}^{n} \ln (1 + 2 x) dx

Answered by mathmax by abdo last updated on 27/Jun/20

A_{n} = \int_{0}^{\infty} {(1 - \frac{x}{n})}^{n} \ln (2 x + 1) dx = \int_{R} {(1 - \frac{x}{n})}^{n} \ln (2 x + 1) χ_{[0, + \infty [} (x) dx

= \int_{R} f_{n} (x) dx with f_{n} (x) = {(1 - \frac{x}{n})}^{n} \ln (2 x + 1)

f_{n} \to^{cs} f (x) = e^{- x} \ln (2 x + 1) and ∣ f_{n} ∣ ⩽ f (x) integrable on [0, + \infty [

tbeorem of convegence dominee give \lim_{n \to + \infty} A_{n} = \int_{0}^{\infty} e^{- x} \ln (2 x + 1) dx = L

by parts L = {[- e^{- x} \ln (2 x + 1)]}_{0}^{+ \infty} + \int_{0}^{\infty} e^{- x} \frac{2}{2 x + 1} dx

= 2 \int_{0}^{\infty} \frac{e^{- x}}{2 x + 1} dx =_{2 x + 1 = t} 2 \int_{1}^{+ \infty} \frac{e^{- (\frac{t - 1}{2})}}{t} \times \frac{dt}{2} = \sqrt{e} \int_{1}^{+ \infty} \frac{e^{- \frac{t}{2}}}{t} dt \dots be continued \dots

Answered by maths mind last updated on 28/Jun/20

e^{- x} ⩾ 1 - x \Leftrightarrow, x > 0

\frac{e^{- x} - 1}{x} ⩾ - 1 \dots t r u e b y m e a n v a l u t t h

f (t) = e^{- t} o v e r [0, x] x > 0 \exists c \in [0, x]

\Rightarrow \frac{e^{- x} - 1}{x} = - e^{- c} ⩾ - 1

\Rightarrow e^{- \frac{x}{n}} ⩾ 1 - \frac{x}{n} \Rightarrow {(1 - \frac{x}{n})}^{n} l n (1 + 2 x) ⩽ e^{- x} l n (1 + 2 x)

\int_{0}^{+ \infty} e^{- x} l n (1 + 2 x) d x < \infty t r u e

\Rightarrow c o n v e r g e n c e t h e o r e m

\Rightarrow \lim_{n \to \infty} \int_{0}^{\infty} {(1 - \frac{x}{n})}^{n} l n (1 + 2 x) d x = \int_{0}^{+ \infty} \lim_{n \to \infty} {(1 - \frac{x}{n})}^{n} l n (1 + 2 x) d x

= \int_{0}^{+ \infty} e^{- x} l n (1 + 2 x) d x

[- e^{- x} l n (1 + 2 x)] + \int_{0}^{+ \infty} \frac{e^{- x}}{1 + 2 x} d x

= \int_{0}^{+ \infty} \frac{e^{- x}}{1 + 2 x} d x = \int_{1}^{+ \infty} \frac{e^{- (\frac{u - 1}{2})}}{2 u} d u

= \frac{\sqrt{e}}{2} \int_{1}^{+ \infty} \frac{e^{- \frac{u}{2}}}{2 u} d u = \frac{\sqrt{e}}{2} \int_{\frac{1}{2}}^{+ \infty} \frac{e^{- t}}{2 t}

= \frac{\sqrt{e}}{4} \int_{\frac{1}{2}}^{+ \infty} \frac{e^{- t}}{t} d t = \frac{\sqrt{e}}{4} E (\frac{1}{2})