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Question Number 62128 by maxmathsup by imad last updated on 15/Jun/19

l e t U_{n} = \int_{0}^{\infty} \frac{c o s (n x)}{{(x^{2} + n^{2})}^{3}} d x w i t h n ⩾ 1

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

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

3) c a l c u l a t e {l i m}_{n \to + \infty} n^{2} U_{n}

4) s t u d y t h e c o n v e r v e n c e o f U_{n}

Commented by maxmathsup by imad last updated on 16/Jun/19

1) U_{n} = \int_{0}^{\infty} \frac{c o s (n x)}{{(x^{2} + n^{2})}^{3}} d x c h a n g e m e n t x = n t g i v e

U_{n} = \int_{0}^{\infty} \frac{c o s (n^{2} t)}{{(n^{2} t^{2} + n^{2})}^{3}} n d t = \frac{1}{n^{5}} \int_{0}^{\infty} \frac{c o s (n^{2} t)}{{(t^{2} + 1)}^{3}} d t \Rightarrow 2 n^{5} U_{n} = \int_{- \infty}^{+ \infty} \frac{c o s (n^{2} t)}{{(t^{2} + 1)}^{3}} d t

= R e (\int_{- \infty}^{+ \infty} \frac{e^{{i n}^{2} t}}{{(t^{2} + 1)}^{3}} d t) l e t c o n s i d e r t h e c o m p l e x f u n c t i o n φ (z) = \frac{e^{{i n}^{2} z}}{{(z^{2} + 1)}^{3}}

w e h a v e φ (z) = \frac{e^{{i n z}^{2}}}{{(z - i)}^{3} {(z + i)}^{3}} t h e p o l e s o f φ a r e \bar{+} i (t r i p l e s) 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)

R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} φ (z)}^{(2)}

= {l i m}_{z \to i} \frac{1}{2} {\frac{e^{{i n}^{2} z}}{{(z + i)}^{3}}}^{(2)} b u t

{\frac{e^{{i n}^{2} z}}{{(z + i)}^{3}}}^{(2)} = {\frac{{i n}^{2} e^{{i n}^{2} z} {(z + i)}^{3} - 3 {(z + i)}^{2} e^{{i n}^{2} z}}{{(z + i)}^{6}}}^{(1)}

= {\frac{({i n}^{2} (z + i) - 3) e^{{i n}^{2} z}}{{(z + i)}^{4}}}^{(1)} = {\frac{({i n}^{2} z - n^{2} - 3) e^{{i n}^{2} z}}{{(z + i)}^{4}}}^{(1)}

= \frac{({i n}^{2} + {i n}^{2} ({i n}^{2} z - n^{2} - 3)) e^{{i n}^{2} z} {(z + i)}^{4} - 4 {(z + i)}^{3} ({i n}^{2} z - n^{2} - 3) e^{{i n}^{2} z}}{{(z + i)}^{8}}

= \frac{{({i n}^{2} - n^{4} z - {i n}^{4} - 3 {i n}^{2}) (z + i) - 4 {i n}^{2} z + 4 n^{2} + 12} e^{{i n}^{2} z}}{{(z + i)}^{5}} \Rightarrow

R e s (φ, i) = \frac{1}{2} \frac{{2 i ({i n}^{2} - n^{4} i - {i n}^{4} - 3 {i n}^{2}) + 4 n^{2} + 4 n^{2} + 12} e^{- n^{2}}}{{(2 i)}^{5}}

= \frac{1}{2^{6}} \frac{{- 2 n^{2} + 2 n^{4} + 2 n^{4} + 6 n^{2} + 8 n^{2} + 12} e^{- n^{2}}}{i} = \frac{1}{i 2^{6}} {12 n^{2} + 4 n^{4} + 12} e^{- n^{2}}

= \frac{4}{i 2^{6}} {3 n^{2} + n^{4} + 3} e^{- n^{2}} = \frac{1}{16 i} {n^{4} + 3 n^{2} + 3} e^{- n^{2}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{16 i} {n^{4} + 3 n^{2} + 3} e^{- n^{2}} = \frac{π}{8} (n^{4} + 3 n^{2} + 3) e^{- n^{2}} = 2 n^{5} U_{n} \Rightarrow

U_{n} = \frac{π}{16 n^{5}} (n^{4} + 3 n^{2} + 3) e^{- n^{2}}

3) n^{2} U_{n} = \frac{π}{16 n^{3}} (n^{4} + 3 n^{2} + 3) e^{- n^{2}} = \frac{π}{16} (n + 3 + \frac{3}{n^{3}}) e^{- n^{2}} \Rightarrow

{l i m}_{n \to + \infty} n^{2} U_{n} = {l i m}_{n \to + \infty} \frac{π}{16} (n + 3) e^{- n^{2}} = 0

Commented by maxmathsup by imad last updated on 16/Jun/19

2) w e h a v e {n U}_{n} = \frac{π}{16 n^{4}} (n^{4} + 3 n^{2} + 3) e^{- n^{2}} \Rightarrow {l i m}_{n \to + \infty} {n U}_{n} =

{l i m}_{n \to + \infty} \frac{π}{16} e^{- n^{2}} = 0

Commented by maxmathsup by imad last updated on 16/Jun/19

4) t h e Q i s s t u d y t h e c o n v e r g e n c e o f Σ U_{n}