find-0-xsin-2x-x-2-4-3-dx-

Question Number 133119 by abdomsup last updated on 19/Feb/21

f i n d \int_{0}^{\infty} \frac{x s i n (2 x)}{{(x^{2} + 4)}^{3}} d x

Answered by mindispower last updated on 19/Feb/21

l e t

f (a) = \int_{0}^{\infty} \frac{x s i n (2 x)}{x^{2} + a} d x, \forall a \in R_{+} - {0}

f (a) = \frac{1}{2} I m \int_{- \infty}^{\infty} \frac{{x e}^{2 i x}}{x^{2} + a} d x \dots . . E

\int_{- \infty}^{\infty} \frac{{x e}^{2 i x}}{x^{2} + a} = 2 i π R e s (\frac{{x e}^{2 i x}}{x^{2} + a}, x = i \sqrt{a})

= 2 i π . \frac{i \sqrt{a} . e^{- 2 \sqrt{a}}}{2 i \sqrt{a}} = i π e^{- 2 \sqrt{a}}

E \Rightarrow f (a) = \frac{π}{2} e^{- 2 \sqrt{a}}

f^{'} (a) = \int_{0}^{\infty} \frac{- x s i n (2 x)}{{(x^{2} + a)}^{2}} d x

f^{″} (a) = 2 \int \frac{x s i n (2 x) d x}{{(x^{2} + a)}^{3}}

\frac{1}{2} f^{″} (4) = \int_{0}^{\infty} \frac{x s i n (2 x)}{{(x^{2} + 4)}^{2}} d x \dots o u r i n t e g r a l

f^{'} (a) = - \frac{π}{2 \sqrt{a}} e^{- 2 \sqrt{a}}, f^{″} (a) = \frac{π}{2 a} e^{- 2 \sqrt{a}} + \frac{π}{4 a \sqrt{a}} e^{- 2 \sqrt{a}}

f^{″} (4) = \frac{5 π}{32} e^{- 4}

Answered by mathmax by abdo last updated on 20/Feb/21

Φ = \int_{0}^{\infty} \frac{xsin (2 x)}{{(x^{2} + 4)}^{3}} dx \Rightarrow Φ =_{x = 2 t} \int_{0}^{\infty} \frac{2 tsin (4 t)}{64 {(t^{2} + 1)}^{3}} (2 dt)

= \frac{1}{16} \int_{0}^{\infty} \frac{tsin (4 t)}{{(t^{2} + 1)}^{3}} dt = \frac{1}{32} \int_{- \infty}^{+ \infty} \frac{tsin (4 t)}{{(t^{2} + 1)}^{3}} dt = \frac{1}{32} Im (\int_{- \infty}^{+ \infty} \frac{{te}^{4 it}}{{(t^{2} + 1)}^{3}} dt)

φ (z) = \frac{{ze}^{4 iz}}{{(z^{2} + 1)}^{3}} \Rightarrow φ (z) = \frac{{ze}^{4 iz}}{{(z - i)}^{3} {(z + i)}^{3}}

\int_{R} φ (z) dz = 2 i π Res (φ, i) and Res (φ, i) = \lim_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} φ (z)}^{(2)}

= \frac{1}{2} \lim_{z \to i} {\frac{{ze}^{4 iz}}{{(z + i)}^{3}}}^{(2)} \Rightarrow 2 Res (φ, i) = \lim_{z \to i} {\frac{(e^{4 iz} + 4 iz e^{4 iz}) {(z + i)}^{3} - 3 {(z + i)}^{2} {ze}^{4 iz}}{{(z + i)}^{6}}}^{(1)}

\lim_{z \to i} {\frac{(1 + 4 iz) e^{4 iz} (z + i) - {3 ze}^{4 iz}}{{(z + i)}^{4}}}^{(1)}

= \lim_{z \to i} {\frac{(1 + 4 iz) (z + i) - 3 z) e^{4 iz}}{{(z + i)}^{4}}}^{(1)}

= \lim_{z \to i} {\frac{(i + {4 iz}^{2} - 3 z) e^{4 iz}}{{(z + i)}^{4}}}^{(1)}

= \lim_{z \to i} \frac{(8 iz - 3) e^{4 iz} + 4 i ({4 iz}^{2} - 3 z + 1) e^{4 iz}) {(z + i)}^{4} - 4 {(z + i)}^{3} ({4 iz}^{2} - 3 z + i) e^{4 iz}}{{(z + i)}^{8}}

\dots . rest to finish the calculus \dots