study-the-convergence-of-U-n-with-U-n-0-cos-nx-x-2-n-2-dx-n-1-

Question Number 64525 by mathmax by abdo last updated on 19/Jul/19

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

U_{n} = \int_{0}^{\infty} \frac{c o s (n x)}{x^{2} + n^{2}} d x (n ⩾ 1)

Commented by mathmax by abdo last updated on 19/Jul/19

w e h a v e 2 U_{n} = \int_{- \infty}^{+ \infty} \frac{c o s (n x)}{x^{2} + n^{2}} =_{x = n t} \int_{- \infty}^{+ \infty} \frac{c o s (n^{2} t)}{n^{2} (1 + t^{2})} n d t

= \frac{1}{n} \int_{- \infty}^{+ \infty} \frac{c o s (n^{2} t)}{t^{2} + 1} d t = \frac{1}{n} R e (\int_{- \infty}^{+ \infty} \frac{e^{{i n}^{2} t}}{t^{2} + 1} d t) l e t φ (z) = \frac{e^{{i n}^{2} z}}{z^{2} + 1}

w e h a v e φ (z) = \frac{e^{{i n z}^{2}}}{(z - i) (z + 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)

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

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

U_{n} = \frac{π}{2 n} e^{- n^{2}} (n ⩾ 1) i t s c l a e a r t h a t \sum_{n = 1}^{\infty} U_{n} ⩽ \frac{π}{2} \sum_{n = 1}^{\infty} e^{- n^{2}}

⩽ \frac{π}{2} \sum_{n = 1}^{\infty} e^{- n} a n d t h i s s e r i e i s c o n v e r g e n t \Rightarrow Σ U_{n} c o n v e r g e s