find-the-value-of-0-cos-xt-t-2-x-2-2-dt-

Question Number 33737 by prof Abdo imad last updated on 23/Apr/18

f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{c o s (x t)}{{(t^{2} + x^{2})}^{2}} d t .

Commented by prof Abdo imad last updated on 23/Apr/18

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

f (x) = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{c o s (x t)}{{(t^{2} + x^{2})}^{2}} d t \Rightarrow

2 f (x) = R e (\int_{- \infty}^{+ \infty} \frac{e^{i x t}}{{(t^{2} + x^{2})}^{2}} d t) l e t i n t r o d u c e

t h e c o m p l e x f u n c t i o n φ (z) = \frac{e^{i x z}}{{(z^{2} + x^{2})}^{2}}

t h e p o l e s o f φ a r e i x, - i x (d o u b l e s)

c a s e 1 x > 0

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i x)

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

= {l i m}_{z \to i x} {(\frac{e^{i x z}}{{(z + i x)}^{2}})}^{^{'}}

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

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

= \frac{i x e^{- x^{2}} (2 i x) - 2 e^{- x^{2}}}{{(2 i x)}^{3}} = \frac{- 2 x^{2} e^{- x^{2}} - 2 e^{- x^{2}}}{- 8 i x^{3}}

= \frac{(x^{2} + 1) e^{- x^{2}}}{4 {i x}^{3}} \Rightarrow 2 f (x) = 2 i π \frac{(x^{2} + 1) e^{- x^{2}}}{4 {i x}^{3}}

\Rightarrow f (x) = \frac{i π}{4} (1 + x^{2}) e^{- x^{2}} .

Commented by prof Abdo imad last updated on 23/Apr/18

e r r o r i n t h e f i n a l l i n e f (x) = \frac{π}{4 x^{3}} (1 + x^{2}) e^{- x^{2}} .

Commented by caravan msup abdo. last updated on 24/Apr/18

c a s e 2 x < 0 \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, - i x)

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

= {l i m}_{x \to - i x} {(\frac{e^{i x z}}{{(z - i x)}^{2}})}^{^{'}}

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

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

= \frac{i x e^{x^{2}} (- 2 i x) - 2 e^{x^{2}}}{{(- 2 i x)}^{3}}

= \frac{2 x^{2} e^{x^{2}} - 2 e^{x^{2}}}{8 {i x}^{3}} = \frac{(x^{2} - 1) e^{x^{2}}}{4 {i x}^{3}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π . \frac{(x^{2} - 1) e^{x^{2}}}{4 {i x}^{3}}

= \frac{π}{2 x^{3}} (x^{2} - 1) e^{x^{2}} \Rightarrow f (x) = \frac{π}{4 x^{3}} (x^{2} - 1) e^{x^{2}} .

Answered by sma3l2996 last updated on 23/Apr/18

\int_{0}^{\infty} \frac{c o s (x t)}{{(t^{2} + x^{2})}^{2}} d t = \frac{1}{2} \int_{- \infty}^{\infty} \frac{c o s (x t)}{{(t^{2} + x^{2})}^{2}} d t = \frac{1}{2} \int_{∣ z ∣\to \infty} R e (\frac{e^{i x t}}{{(t^{2} + x^{2})}^{2}}) d t

l e t f (t) = \frac{e^{i x t}}{{(t^{2} + x^{2})}^{2}} = \frac{e^{i x t}}{{(t + i x)}^{2} {(t - i x)}^{2}}

i f x > 0

s o \int_{0}^{\infty} f (t) d t = i π R e s (f; i x)

R e (f; i x) = \underset{t \to i x}{l i m} \frac{d}{d t} (\frac{e^{i x t}}{{(t + i x)}^{2}}) = \underset{t \to i x}{l i m} \frac{e^{i x t} (i x t - x^{2} - 2)}{{(t + i x)}^{3}}

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

S o \int_{0}^{+ \infty} \frac{c o s (x t)}{{(t^{2} + x^{2})}^{2}} d t = \frac{π e^{- x^{2}} (x^{2} + 1)}{4 x^{3}}