calculate-f-t-0-cos-tx-1-tx-2-2-dx-with-t-0-2-find-the-values-of-0-cos-2x-1-2x-2-2-dx-and-0-cosx-2-x-2-2-dx-

Question Number 38470 by maxmathsup by imad last updated on 25/Jun/18

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

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

Commented by math khazana by abdo last updated on 26/Jun/18

t > 0

Commented by math khazana by abdo last updated on 27/Jun/18

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

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

φ (z) = \frac{e^{i t z}}{{({t z}^{2} + 1)}^{2}} w e h a v e

φ (z) = \frac{e^{i t z}}{t^{2} {(z^{2} + \frac{1}{t})}^{2}} = \frac{e^{i t z}}{t^{2} {(z - \frac{i}{\sqrt{t}})}^{2} {(z + \frac{i}{\sqrt{t}})}^{2}}

t h e p o l e s o f φ a r e \frac{i}{\sqrt{t}} a n d \frac{- i}{\sqrt{t}} (d o u b l e s)

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, \frac{i}{\sqrt{t}}) b u t

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

= {l i m}_{z \to \frac{i}{\sqrt{t}}} \frac{1}{t^{2}} {\frac{e^{i t z}}{{(z + \frac{i}{\sqrt{t}})}^{2}}}^{(1)}

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

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

= \frac{1}{t^{2}} \frac{i t e^{i t \frac{i}{\sqrt{t}}} (\frac{2 i}{\sqrt{t}}) - 2 e^{i t (\frac{i}{\sqrt{t}})}}{{(\frac{2 i}{\sqrt{t}})}^{3}} = \frac{1}{t^{2}} \frac{- 2 \sqrt{t} e^{- \sqrt{t}} - 2 e^{- \sqrt{t}}}{\frac{- 8 i}{t \sqrt{t}}}

= \frac{t \sqrt{t}}{t^{2}} \frac{(1 + \sqrt{t}) e^{- \sqrt{t}}}{4 i} = \frac{\sqrt{t}}{4 t i} (1 + \sqrt{t}) e^{- \sqrt{t}}

= \frac{(t + \sqrt{t}) e^{- \sqrt{t}}}{4 t i} \Rightarrow

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

2 f (x) = R e (\int_{- \infty}^{+ \infty} φ (z) d z) \Rightarrow

f (x) = \frac{π}{4 t} (t + \sqrt{t}) e^{- \sqrt{t}} (t > 0)

2) \int_{0}^{\infty} \frac{c o s (2 x)}{{(1 + 2 x^{2})}^{2}} d x = f (2) = \frac{π}{8} (2 + \sqrt{2}) e^{- \sqrt{2}}

Commented by math khazana by abdo last updated on 27/Jun/18

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

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

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

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

i \sqrt{2} a n d - i \sqrt{2} (d o u b l e s)

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

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

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

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

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

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

= \frac{- 2 \sqrt{2} e^{- \sqrt{2}} - 2 e^{- \sqrt{2}}}{- 8 i (2 \sqrt{2})} = \frac{(1 + \sqrt{2}) e^{- \sqrt{2}}}{8 i \sqrt{2}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{(1 + \sqrt{2}) e^{- \sqrt{2}}}{8 i \sqrt{2}}

= \frac{π}{4 \sqrt{2}} (1 + \sqrt{2}) e^{- \sqrt{2}}

2 I = R e (\int_{- \infty}^{+ \infty} φ (z) d z) \Rightarrow

I = \frac{π}{8 \sqrt{2}} (1 + \sqrt{2}) e^{- \sqrt{2}} .

Commented by abdo.msup.com last updated on 27/Jun/18

f (t) = \frac{π}{4 t} (t + \sqrt{t}) e^{- \sqrt{t}}