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Question Number 37285 by math khazana by abdo last updated on 11/Jun/18

l e t A_{n} = \int_{0}^{\infty} e^{- {n x}^{2}} s i n (\frac{x}{n}) d x w i t h n i n t e g r n o t 0

1) c a l c u l a t e A_{n}

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

Commented by math khazana by abdo last updated on 14/Jun/18

A_{n} = \int_{0}^{+ \infty} e^{- {n x}^{2}} c o s (\frac{x}{n}) d x .

Commented by prof Abdo imad last updated on 16/Jun/18

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

A_{n} = \int_{0}^{\infty} e^{- {n n}^{2} t^{2}} c o s (t) n d t

= n \int_{0}^{\infty} e^{- n^{3} t^{2}} c o s t d t

= \frac{n}{2} \int_{- \infty}^{+ \infty} e^{- n^{3} t^{2} + i t} d t

\frac{2}{n} A_{n} = \int_{- \infty}^{+ \infty} e^{- {{(n \sqrt{n} t)}^{2} - i t}} d t

= \int_{- \infty}^{+ \infty} e^{- {{(n \sqrt{n} t)}^{2} - 2 \frac{i}{n \sqrt{n}} (n \sqrt{n} t) + {(\frac{i}{n \sqrt{n}})}^{2} - {(\frac{i}{n \sqrt{n}})}^{2}}} d t

= \int_{- \infty}^{+ \infty} e^{- {(n \sqrt{n} t - \frac{i}{n \sqrt{n}})}^{2} - \frac{1}{n^{3}}} d t

=_{n \sqrt{n} t - \frac{i}{n \sqrt{n}} = u} e^{- \frac{1}{n^{3}}} \int_{- \infty}^{+ \infty} e^{- u^{2}} d u

= \sqrt{π} e^{- \frac{1}{n^{3}}} \Rightarrow

A_{n} = \frac{n \sqrt{π}}{2} e^{- \frac{1}{n^{3}}} .

Commented by math khazana by abdo last updated on 17/Jun/18

2) i t s c l e a r t h a t {l i m}_{n \to + \infty} A_{n} = + \infty .