we-give-0-e-x-ln-x-dx-1-calculate-f-a-0-e-ax-ln-x-dx-with-a-gt-0-2-let-u-n-0-e-nx-ln-x-n-dx-find-lim-n-u-n-

Question Number 38198 by maxmathsup by imad last updated on 22/Jun/18

w e g i v e \int_{0}^{\infty} e^{- x} l n (x) d x = - γ

1) c a l c u l a t e f (a) = \int_{0}^{\infty} e^{- a x} l n (x) d x w i t h a > 0

2) l e t u_{n} = \int_{0}^{\infty} e^{- n x} l n (\frac{x}{n}) d x f i n d {l i m}_{n \to + \infty} u_{n}

Commented by abdo.msup.com last updated on 24/Jun/18

1) c h a n g e m e n t a x = t g i v e

f (a) = \int_{0}^{\infty} e^{- t} l n (\frac{t}{a}) \frac{d t}{a}

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

= \frac{1}{a} {γ - l n (a) {[- e^{- t}]}_{0}^{+ \infty}}

= \frac{γ + l n (a)}{a}

f (a) = \frac{γ + l n (a)}{a} w i t h a > 0

Commented by abdo.msup.com last updated on 24/Jun/18

2) c h a n g e m e n t n x = t g i v e

u_{n} = \int_{0}^{\infty} e^{- t} l n (\frac{t}{n^{2}}) \frac{d t}{n}

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

= \frac{1}{n} {γ + 2 l n (n)}

{l i m}_{n \to + \infty} u_{n} = {l i m}_{n \to + \infty} (\frac{γ}{n} + 2 \frac{l n (n)}{n}) = 0

Commented by math khazana by abdo last updated on 24/Jun/18

γ i s t h e c o s t a n t o f E u l e r .

Commented by math khazana by abdo last updated on 24/Jun/18

f (a) = \frac{- γ + l n (a)}{a}

Commented by math khazana by abdo last updated on 24/Jun/18

u_{n} = \frac{- γ + 2 l n (n)}{n}