let-A-n-0-x-sin-nx-x-2-n-2-2-dx-with-n-integr-natural-not-0-1-find-the-value-of-A-n-2-study-the-convergence-of-A-n-

Question Number 53114 by maxmathsup by imad last updated on 17/Jan/19

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

1) f i n d t h e v a l u e o f A_{n}

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

Commented by maxmathsup by imad last updated on 18/Jan/19

1) c h a n g e m e n t x = n t g i v e A_{n} = \int_{0}^{\infty} \frac{n t s i n (n^{2} t)}{n^{4} {(t^{2} + 1)}^{2}} n d t = \frac{1}{n^{2}} \int_{0}^{\infty} \frac{t s i n (n^{2} t)}{{(t^{2} + 1)}^{2}} d t \Rightarrow

2 n^{2} A_{n} = \int_{- \infty}^{+ \infty} \frac{t s i n (n^{2} t)}{{(t^{2} + 1)}^{2}} d t = I m (\int_{- \infty}^{+ \infty} \frac{t e^{{i n}^{2} t}}{{(t^{2} + 1)}^{2}} 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{z e^{{i n}^{2} z}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{z e^{{i n}^{2} z}}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p o l e s o f φ a r e i a n d - i

(d o u b 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}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}

= {l i m}_{z \to i} {\frac{z e^{{i n}^{2} z}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{(e^{{i n}^{2} z} + {i n}^{2} z e^{{i n}^{2} z}) {(z + i)}^{2} - 2 (z + i) z e^{{i n}^{2} z}}{{(z + i)}^{4}}

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

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

2 n^{2} A_{n} = \frac{π n^{2}}{2} e^{- n^{2}} \Rightarrow A_{n} = \frac{π}{4} e^{- n^{2}} .

w e h a v e \sum_{n = 0}^{\infty} A_{n} = \frac{π}{4} \sum_{n = 0}^{\infty} e^{- n^{2}} b u t n^{2} ⩾ n \Rightarrow - n^{2} ⩽ - n \Rightarrow e^{- n^{2}} ⩽ e^{- n} \Rightarrow

\sum_{n = 0}^{\infty} A_{n} ⩽ \frac{π}{4} \sum_{n = 0}^{\infty} {(\frac{1}{e})}^{n} = \frac{π}{4} \frac{1}{1 - \frac{1}{e}} = \frac{π}{4} \frac{e}{e - 1} \Rightarrow

0 < \sum_{n = 0}^{\infty} A_{n} ⩽ \frac{π e}{4 (e - 1)} .