let-I-n-0-1-1-x-1-1-x-n-n-dx-and-J-n-1-n-1-x-1-x-n-n-dx-n-integr-not-0-1-prove-that-lim-I-n-0-1-1-e-x-x-dx-lim-J-n-0-1-e-1-x-x-dx-n-2-

Question Number 33342 by prof Abdo imad last updated on 14/Apr/18

l e t I_{n} = \int_{0}^{1} \frac{1}{x} (1 - {(1 - \frac{x}{n})}^{n}) d x a n d

J_{n} = \int_{1}^{n} \frac{1}{x} {(1 - \frac{x}{n})}^{n} d x, n i n t e g r n o t 0

1) p r o v e t h a t {l i m}_{} I_{n} = \int_{0}^{1} \frac{1 - e^{- x}}{x} d x

l i m J_{n} = \int_{0}^{1} \frac{e^{- \frac{1}{x}}}{x} d x (n \to \infty)

2) p r o v e t h a t \forall n \in N^{★} I_{n} - J_{n} = \sum_{k = 1}^{n} \frac{1}{k} - l n (n)

3) p r o v e t h a t \int_{0}^{1} \frac{1 - e^{- x} - e^{- \frac{1}{x}}}{x} = γ

Commented by prof Abdo imad last updated on 24/Apr/18

I_{n} = \int_{R} \frac{1}{x} (1 - {(1 - \frac{x}{n})}^{n}) χ_{] \bar{01}]} (x) d x l e t p u t

f_{n} (x) = \frac{1}{x} (1 - {(1 - \frac{x}{n})}^{n}) χ_{] 0, 1]} (x)

f_{n} (x)^{c . s} \to \frac{1 - e^{- x}}{x} χ_{] 0, 1]} (x) (n \to + \infty) a n d

f_{n} (x) ⩽ \frac{1 - e^{- x}}{x} \forall x \in] 0, 1] c o n v . d o m i n e e \Rightarrow

\int_{R} f_{n} (x) d x_{n \to + \infty} \to \int_{0}^{1} \frac{1 - e^{- x}}{x} d x \Rightarrow

{l i m}_{n \to + \infty} I_{n} = \int_{0}^{1} \frac{1 - e^{- x}}{x} d x .

Commented by prof Abdo imad last updated on 24/Apr/18

c h a n g e m e n t x = \frac{1}{t} g i v e

J_{n} = - \int_{1}^{\frac{1}{n}} t {(1 - \frac{1}{n t})}^{n} \frac{d t}{t^{2}} = \int_{\frac{1}{n}}^{1} \frac{1}{t} {(1 - \frac{1}{n t})}^{n} d t \Rightarrow

J_{n} \to \int_{0}^{1} \frac{e^{- \frac{1}{t}}}{t} d t (n \to + \infty) s o

{l i m}_{n \to + \infty} J_{n} = \int_{0}^{1} \frac{e^{- \frac{1}{t}}}{t} d t .

Commented by math khazana by abdo last updated on 25/Apr/18

c h a n g e m e n t \frac{x}{n} = t g i v e

J_{n} = \int_{\frac{1}{n}}^{1} \frac{1}{n t} {(1 - t)}^{n} n d t = \int_{\frac{1}{n}}^{1} \frac{{(1 - t)}^{n}}{t} d t \Rightarrow

I_{n} - J_{n} = \int_{0}^{1} \frac{1}{x} (1 - {(1 - \frac{x}{n})}^{n}) d x + \int_{1}^{\frac{1}{n}} \frac{{(1 - x)}^{n}}{x} d x

= \int_{0}^{\frac{1}{n}} \frac{1}{x} (1 - {(1 - \frac{x}{n})}^{n} + {(1 - x)}^{n}) d x

= \int_{0}^{\frac{1}{n}} \frac{1}{x} (1 + \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} x^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} \frac{x^{k}}{n^{k}}) d x

= \int_{0}^{\frac{1}{n}} \frac{1}{x} (\sum_{k = 0}^{n} C_{n}^{k} {(- 1)}^{k} x^{k} - \sum_{k = 1}^{n} C_{n}^{k} {(- 1)}^{k} \frac{x^{k}}{n^{k}}) d x

= \int_{0}^{\frac{1}{n}} \frac{1}{x} (1 + \sum_{k = 1}^{n} C_{n}^{k} {(- 1)}^{k} (1 + \frac{1}{n^{k}}) x^{k}) d x \dots b e

c o n t i n u e d \dots

Commented by math khazana by abdo last updated on 25/Apr/18

3) w e h a v e I_{n} - J_{n} = \sum_{k = 1}^{n} \frac{1}{k} - l n (n) \Rightarrow

{l i m}_{n \to + \infty} (I_{n} - J_{n}) = {l i m}_{n \to + \infty} (\sum_{k = 1}^{n} \frac{1}{k} - l n (n)) = γ

\Rightarrow \int_{0}^{1} \frac{1 - e^{- x}}{x} d x - \int_{0}^{1} \frac{e^{- \frac{1}{x}}}{x} d x = γ \Rightarrow

\int_{0}^{1} \frac{1 - e^{- x} - e^{- \frac{1}{x}}}{x} = γ .