calculate-lim-n-0-dx-x-n-e-x-

Question Number 33694 by math khazana by abdo last updated on 22/Apr/18

c a l c u l a t e {l i m}_{n \to + \infty} \int_{0}^{\infty} \frac{d x}{x^{n} + e^{x}} .

Commented by math khazana by abdo last updated on 29/Apr/18

l e t p u t A_{n} = \int_{0}^{n} \frac{d x}{x^{n} + e^{x}}

A_{n} = \int_{0}^{1} \frac{d x}{x^{n} + e^{x}} + \int_{1}^{n} \frac{d x}{x^{n} + e^{x}} b u t

\int_{0}^{1} \frac{d x}{x^{n} + e^{x}} \to \int_{0}^{1} e^{- x} d x = {[- e^{- x}]}_{0}^{1} = 1 - \frac{1}{e}

c h . x^{n} = t g i v e

\int_{1}^{n} \frac{d x}{x^{n} + e^{x}} = \int_{1}^{n^{n}} \frac{1}{t + e^{t^{\frac{1}{n}}}} \frac{1}{n} t^{\frac{1}{n} - 1} d t

= \frac{1}{n} \int_{1}^{n^{n}} \frac{t^{\frac{1}{n}}}{t^{2} + t e^{t^{\frac{1}{n}}}} d t ⩽ \frac{1}{n} \int_{1}^{+ \infty} \frac{t^{\frac{1}{n}}}{t^{2} + t e^{t^{\frac{1}{n}}}} d t

⩽ \frac{1}{n} \int_{1}^{+ \infty} \frac{t^{\frac{1}{n}}}{t^{2}} d t \to 0 (n \to + \infty) s o

{l i m}_{n \to + \infty} A_{n} = 1 - \frac{1}{e} .