calculate-lim-n-0-n-e-nx-1-nx-2-dx-

Question Number 55229 by maxmathsup by imad last updated on 19/Feb/19

c a l c u l a t e {l i m}_{n \to + \infty} \int_{0}^{n} \frac{e^{n x}}{1 + {n x}^{2}} d x .

Commented by maxmathsup by imad last updated on 25/Feb/19

l e t I_{n} = \int_{0}^{n} \frac{e^{n x}}{1 + {n x}^{2}} d x w e h a v e 1 + {n x}^{2} ⩽ n (1 + x^{2}) \Rightarrow \frac{1}{1 + {n x}^{2}} ⩾ \frac{1}{n (1 + x^{2})} \Rightarrow

\frac{e^{n x}}{1 + {n x}^{2}} ⩾ \frac{e^{n x}}{n (1 + x^{2})} \Rightarrow I_{n} ⩾ \frac{1}{n} \int_{0}^{n} \frac{e^{n x}}{1 + x^{2}} d x b u t

e^{n x} = \sum_{p = 0}^{\infty} \frac{{(n x)}^{p}}{p!} \Rightarrow \frac{e^{n x}}{1 + x^{2}} = \sum_{p = 0}^{\infty} \frac{n^{p} x^{p}}{p! (1 + x^{2})} \Rightarrow \frac{1}{n} \int_{0}^{n} \frac{e^{n x}}{1 + x^{2}} d x

= \frac{1}{n} \sum_{p = 0}^{\infty} n^{p} \int_{0}^{n} \frac{x^{p}}{p! (1 + x^{2})} d x = \int_{0}^{n} \frac{d x}{1 + x^{2}} + \sum_{p = 1}^{\infty} n^{p - 1} \int_{0}^{n} \frac{x^{p}}{p! (1 + x^{2})} d x \Rightarrow

{l i m}_{n \to + \infty} \frac{1}{n} \int_{0}^{n} \frac{e^{n x}}{1 + x^{2}} d x = + \infty \Rightarrow {l i m}_{n \to + \infty} I_{n} = + \infty .