nice-mathematics-please-evaluate-x-2-4-x-2-4-sin-2x-x-dx-m-n-1970-

Question Number 117121 by mnjuly1970 last updated on 09/Oct/20

\dots n i c e m a t h e m a t i c s . .

p l e a s e e v a l u a t e \dots

Ω = \int_{- \infty}^{+ \infty} (\frac{x^{2} - 4}{x^{2} + 4} * \frac{s i n (2 x)}{x}) d x = ? ? ?

m . n . 1970

Answered by Bird last updated on 10/Oct/20

I = \int_{- \infty}^{+ \infty} \frac{(x^{2} - 4) s i n (2 x)}{x (x^{2} + 4)} d x \Rightarrow

I = I m (\int_{- \infty}^{+ \infty} \frac{(x^{2} - 4) e^{2 i x}}{x (x^{2} + 4)} d x)

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^{2} - 4) e^{2 i z}}{z (z^{2} + 4)} \Rightarrow

φ (z) = \frac{(z^{2} - 4) e^{2 i z}}{z (z - 2 i) (z + 2 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, o) + R e s (φ, 2 i)}

R e s (φ, o) = {l i m}_{z \to 0} z φ (z)

= {l i m}_{z \to 0} \frac{- 4}{4} = - 1

R e s (φ, 2 i) = {l i m}_{z \to 2 i} (z - 2 i) φ (z)

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

= \frac{- 8 e^{- 4}}{(2 i) (4 i)} = e^{- 4} \Rightarrow

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

I = 2 π {e^{- 4} - 1}

Commented by AbduraufKodiriy last updated on 10/Oct/20

T h i s s o l u t i o n i s w r o n g . B e c a u s e, a c c o r d i n g t o

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i s c o r r e c t .

S o r r y f o r m y m i s t a k e s i n E n g l i s h .

Answered by AbduraufKodiriy last updated on 09/Oct/20

Ω = \int_{- \infty}^{\infty} \frac{x^{2} - 4}{x^{2} + 4} \cdot \frac{s i n 2 x}{x} d x = 2 \int_{0}^{\infty} (1 - \frac{8}{x^{2} + 4}) \frac{s i n 2 x}{x} d x =

= 2 \int_{0}^{\infty} \frac{s i n 2 x}{x} d x - 16 \int_{0}^{\infty} \frac{s i n 2 x}{x (x^{2} + 4)} d x = π - 16 \int_{0}^{\infty} \frac{s i n 2 x}{x (x^{2} + 4)} d x;

ρ = \int_{0}^{\infty} \frac{s i n 2 x}{x (x^{2} + 4)} d x; ρ (t) = \int_{0}^{\infty} \frac{s i n (t x)}{x (x^{2} + 4)} d x;

L (ρ (t)) = L (\int_{0}^{\infty} \frac{s i n (t x)}{x (x^{2} + 4)} d x) = \int_{0}^{\infty} e^{- s t} \int_{0}^{\infty} \frac{s i n (t x)}{x (x^{2} + 4)} d x d t =

= \int_{0}^{\infty} \frac{1}{x (x^{2} + 4)} \int_{0}^{\infty} e^{- s t} s i n (t x) d t d x;

I = \int_{0}^{\infty} e^{- s t} s i n (t x) d t = - \frac{1}{s} e^{- s t} s i n (t x) ∣_{0}^{\infty} - \frac{x}{s^{2}} e^{- s t} c o s (t x) ∣_{0}^{\infty} - \frac{x^{2}}{s^{2}} I \Rightarrow

\Rightarrow (1 + \frac{x^{2}}{s^{2}}) I = \frac{x}{s^{2}} \Rightarrow I = \frac{x}{s^{2} + x^{2}}

L (ρ (t)) = \int_{0}^{\infty} \frac{1}{(x^{2} + 4) (x^{2} + s^{2})} d x = \frac{1}{s^{2} - 4} \int (\frac{1}{x^{2} + 4} - \frac{1}{x^{2} + s^{2}}) d x =

= \frac{1}{s^{2} - 4} (\frac{1}{2} a r c t a n (\frac{x}{2}) - \frac{1}{s} a r c t a n (\frac{x}{s})) ∣_{0}^{\infty} =

= \frac{1}{s^{2} - 4} \cdot (\frac{1}{2} - \frac{1}{s}) \frac{π}{2} = \frac{π}{4} \cdot \frac{1}{s (s + 2)} = \frac{π}{8} (\frac{1}{s} - \frac{1}{s + 2})

L^{- 1} (L (ρ (t))) = L^{- 1} (\frac{π}{8} \cdot (\frac{1}{s} - \frac{1}{s + 2})) \Rightarrow

\Rightarrow ρ (t) = \frac{π}{8} (1 - e^{- 2 t}); ρ = ρ (t = 2) \Rightarrow ρ = \frac{π}{8} (1 - e^{- 4})

Ω = π - 16 ρ = π - 2 π (1 - e^{- 4}) = - π (1 - 2 e^{- 4})

Commented by mnjuly1970 last updated on 09/Oct/20

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