let-f-x-0-cos-xt-xt-2-i-2-dx-with-x-from-R-and-x-0-1-find-a-explicit-form-of-f-x-2-extract-A-Re-f-x-and-B-Im-f-x-and-find-its-values-3-calculate-0-cos-2

Question Number 55760 by maxmathsup by imad last updated on 03/Mar/19

l e t f (x) = \int_{0}^{\infty} \frac{c o s (x t)}{{({x t}^{2} + i)}^{2}} d x w i t h x f r o m R a n d x \neq 0

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

2) e x t r a c t A = R e (f (x)) a n d B = I m (f (x)) a n d f i n d i t s v a l u e s .

3) c a l c u l a t e \int_{0}^{\infty} \frac{c o s (2 t)}{{(2 t^{2} + i)}^{2}} d t

4) l e t U_{n} = \int_{0}^{\infty} \frac{c o s (n t)}{{({n t}^{2} + i)}^{2}} d t . c a l c u l a t e {l i m}_{n \to + \infty} u_{n}

a n d 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 maxmathsup by imad last updated on 04/Mar/19

i^{2} = - 1

Commented by maxmathsup by imad last updated on 04/Mar/19

f (x) = \int_{0}^{\infty} \frac{c o s (x t)}{{({x t}^{2} + i)}^{2}} d t

Commented by maxmathsup by imad last updated on 10/Mar/19

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

= \frac{1}{x} \int_{0}^{\infty} \frac{c o s u}{{(\frac{u^{2}}{x} + i)}^{2}} = x \int_{0}^{\infty} \frac{c o s u}{{(u^{2} + i x)}^{2}} d u = x \int_{0}^{\infty} \frac{{(u^{2} - i x)}^{2} c o s u}{{(u^{4} + x^{2})}^{2}} d u

= x \int_{0}^{\infty} \frac{(u^{2} - 2 i x u^{2} - x^{2}) c o s u}{{(u^{4} + x^{2})}^{2}} d u

= x \int_{0}^{\infty} \frac{(u^{2} - x^{2}) c o s u}{{(u^{4} + x^{2})}^{2}} d u - 2 {i x}^{2} \int_{0}^{\infty} \frac{u^{2} c o s u}{{(u^{4} + x^{2})}^{2}} \Rightarrow R e (f (x)) = x \int_{0}^{\infty} \frac{(u^{2} - x^{2}) c o s u}{{(u^{4} + x^{2})}^{2}} d u = I

a n d I m (f (x)) = - 2 x^{2} \int_{0}^{\infty} \frac{u^{2} c o s u}{(u^{4} + x^{2})} d u l e t f i n d I w e h a v e

2 I = x \int_{- \infty}^{+ \infty} \frac{(u^{2} - x^{2}) c o s u}{{(u^{4} + x^{2})}^{2}} d u = x R e (\int_{- \infty}^{+ \infty} \frac{(u^{2} - x^{2}) e^{i u}}{{(u^{4} + x^{2})}^{2}} d u)

l e t φ (z) = \frac{(z^{2} - x^{2}) e^{i z}}{{(z^{4} + x^{2})}^{2}} \Rightarrow φ (z) = \frac{(z^{2} - x^{2}) e^{i z}}{{(z^{2} - i x)}^{2} {(z^{2} + i x)}^{2}}

= \frac{(z^{2} - x^{2}) e^{i z}}{{(z - \sqrt{x} e^{i \frac{π}{4}})}^{2} {(z + \sqrt{x} e^{\frac{i π}{4}})}^{2} {(z - \sqrt{x} e^{- \frac{i π}{4}})}^{2} {(z + \sqrt{x} e^{- \frac{i π}{4}})}^{2}} s o t h e p o l e s o f φ a r e

\bar{+} \sqrt{x} e^{\frac{i π}{4}} a n d \bar{+} \sqrt{x} e^{- \frac{i π}{4}} (d o u b l e p o 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, \sqrt{x} e^{\frac{i π}{4}}) + R e s (φ, - \sqrt{x} e^{- \frac{i π}{4}})}

R e s (φ, \sqrt{x} e^{\frac{i π}{4}}) = {l i m}_{z \to \sqrt{x} e^{\frac{i π}{4}}} \frac{1}{(2 - 1)!} {{(z - \sqrt{x} e^{\frac{i π}{4}})}^{2} φ (z)}^{(1)}

= {l i m}_{z \to \sqrt{x} e^{\frac{i π}{4}}} {\frac{(z^{2} - x^{2}) e^{i z}}{{(z + \sqrt{x} e^{\frac{i π}{4}})}^{2} {(z^{2} + i x)}^{2}}}^{(1)} \dots . b e c o n t i n u e d \dots

m i m