let-U-n-0-e-n-x-2-x-2-3-dx-1-calculate-U-n-interms-of-n-2-find-lim-n-n-U-n-3-determine-nature-of-the-serie-U-n-

Question Number 60263 by maxmathsup by imad last updated on 19/May/19

l e t U_{n} = \int_{0}^{\infty} \frac{e^{- n [x^{2}]}}{x^{2} + 3} d x

1) c a l c u l a t e U_{n} i n t e r m s o f n

2) f i n d {l i m}_{n \to + \infty} n U_{n}

3) d e t e r m i n e n a t u r e o f t h e s e r i e Σ U_{n}

Commented by maxmathsup by imad last updated on 20/May/19

1) w e h a v e 2 U_{n} = \int_{- \infty}^{+ \infty} \frac{e^{- n [x^{2}]}}{x^{2} + 3} d x l e t φ (z) = \frac{e^{- n [z^{2}]}}{z^{2} + 3} w e h a v e

φ (z) = \frac{e^{- n [z^{2}]}}{(z - i \sqrt{3}) (z + i \sqrt{3})} s o t h e p o l e s o f φ a r e \bar{+} i \sqrt{3} r e s i d u s t h e o r e m g i v e

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i \sqrt{3} 3

R e s (φ, i) = {l i m}_{z \to i \sqrt{3}} (z - i \sqrt{3}) φ (z) = {l i m}_{z \to i} \frac{e^{- n [z^{2}]}}{z + i \sqrt{3}} = \frac{e^{- n [- 3]}}{2 i \sqrt{3}}

= \frac{e^{3 n}}{2 i \sqrt{3}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{e^{3 n}}{2 i \sqrt{3}} = \frac{π}{\sqrt{3}} e^{3 n} \Rightarrow U_{n} = \frac{π}{2 \sqrt{3}} e^{3 n} .

2) {l i m}_{n \to + \infty} {n U}_{n} = + \infty

3) i t s c l e a r t h a t Σ U_{n} d i v e r g e s . .