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Question Number 40984 by prof Abdo imad last updated on 30/Jul/18

l e t f (x) = \int_{0}^{\infty} {t e}^{- t^{2}} a r c t a n (x t) d t

1) f i n d a s i m p l e f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{\infty} {t e}^{- t^{2}} a r c t a n t d t a n d

\int_{0}^{\infty} t e^{- t^{2}} a r c t a n (2 t) d t

3) l e t u_{n} = \int_{0}^{\infty} t e^{- t^{2}} a r c t a n (n t) d t

f i n d {l i m}_{n \to + \infty} u_{n}

s t u d y t h e c o n v e r g e n c e o f Σ u_{n}

Commented by math khazana by abdo last updated on 02/Aug/18

1) b y p a r t s f (x) = {[- \frac{1}{2} e^{- t^{2}} a r c t a n (x t)]}_{t = 0}^{\infty}

- \int_{0}^{\infty} - \frac{1}{2} e^{- t^{2}} \frac{x}{1 + x^{2} t^{2}} d t

= \frac{x}{2} \int_{0}^{\infty} \frac{e^{- t^{2}}}{x^{2} t^{2} + 1} d t =_{x t = u} \frac{x}{2} \int_{0}^{\infty} \frac{e^{- \frac{u^{2}}{x^{2}}}}{1 + u^{2}} \frac{d u}{x}

= \frac{1}{2} \int_{0}^{\infty} \frac{e^{- \frac{u^{2}}{x^{2}}}}{1 + u^{2}} d u = \frac{1}{4} \int_{- \infty}^{+ \infty} \frac{e^{- \frac{u^{2}}{x^{2}}}}{1 + u^{2}} d u l e t

φ (z) = \frac{e^{- \frac{u^{2}}{x^{2}}}}{1 + u^{2}} t h e p o l e s o f φ a r e i a n d - i

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) = 2 i π \frac{e^{\frac{1}{x^{2}}}}{2 i} = π e^{\frac{1}{x^{2}}} \Rightarrow

f (x) = \frac{π}{4} e^{\frac{1}{x^{2}}} w i t h x \neq 0

2) \int_{0}^{\infty} t e^{- t^{2}} a r c t a n (t) d t = f (1) = \frac{π e}{4}

3) w e h a v e u_{n} = f (n) = \frac{π}{4} e^{\frac{1}{n^{2}}} \Rightarrow

{l i m}_{n \to + \infty} u_{n} = \frac{π}{4}

\sum_{n = 1}^{\infty} u_{n} = \frac{π}{4} \sum_{n = 1}^{\infty} e^{\frac{1}{n^{2}}} b u t e^{\frac{1}{n^{2}}} \sim 1 + \frac{1}{n^{2}} \Rightarrow

\frac{π}{4} \sum_{n = 1}^{\infty} e^{\frac{1}{n^{2}}} \sim \frac{π}{4} \sum_{n = 1}^{\infty} (1 + \frac{1}{n^{2}}) = + \infty \Rightarrow

Σ u_{n} d i v e r g e s .