calculate-0-x-2-1-2x-2-3-3-dx-

Question Number 36419 by abdo.msup.com last updated on 01/Jun/18

c a l c u l a t e \int_{0}^{\infty} \frac{x^{2} - 1}{{(2 x^{2} + 3)}^{3}} d x

Commented by abdo.msup.com last updated on 04/Jun/18

w e h a v e I = \int_{0}^{\infty} \frac{x^{2} - 1}{8 {(x^{2} + \frac{3}{2})}^{3}} d x

8 I = \int_{0}^{\infty} \frac{x^{2} - 1}{{(x^{2} + \frac{3}{2})}^{3}} d x

16 I = \int_{- \infty}^{+ \infty} \frac{x^{2} - 1}{{(x^{2} + \frac{3}{2})}^{3}} 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} - 1}{{(z^{2} + \frac{3}{2})}^{3}} w e h a v e

φ (z) = \frac{z^{2} - 1}{{(z - i \sqrt{\frac{3}{2}})}^{3} {(z + i \sqrt{\frac{3}{2}})}^{3}}

t h e ? r o o t s o f φ a r e i \sqrt{\frac{3}{2}} a n d - i \sqrt{\frac{3}{2}}

(t r i p l e s) s o

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i \sqrt{\frac{3}{2}})

R e s (φ, i \sqrt{\frac{3}{2}}) = {l i m}_{z \to i \sqrt{\frac{3}{2}}} \frac{1}{(3 - 1)!} {{(z - i \sqrt{\frac{3}{2}})}^{3} φ (z)}^{(2)}

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

= {l i m}_{z \to i \sqrt{\frac{3}{2}}} \frac{1}{2} {\frac{2 z {(z + i \sqrt{\frac{3}{2}})}^{3} - 3 {(z + i \sqrt{\frac{3}{2}})}^{2} (z^{2} - 1)}{{(z + i \sqrt{\frac{3}{2}})}^{6}}}^{(1)}

= {l i m}_{z \to i \sqrt{\frac{3}{2}}} \frac{1}{2} {\frac{2 z (z + i \sqrt{\frac{3}{2}}) - 3 (z^{2} - 1)}{{(z + i \sqrt{\frac{3}{2}})}^{4}}}^{(1)}

\dots

Commented by abdo.msup.com last updated on 04/Jun/18

w e h a v e {\frac{2 z (z + i \sqrt{\frac{3}{2}}) - 3 (z^{2} - 1)}{{(z + i \sqrt{\frac{3}{2}})}^{4}}}^{(1)}

= {\frac{- z^{2} + 2 i \sqrt{\frac{3}{2}} z + 3}{{(z + i \sqrt{\frac{3}{2}})}^{4}}}^{(1)}

= \frac{(- 2 z + 2 i \sqrt{\frac{3}{2}}) {(z + i \sqrt{\frac{3}{2}})}^{4} - 4 {(z + i \sqrt{\frac{3}{2}})}^{3} (- z^{2} + 2 i \sqrt{\frac{3}{2}} z + 3)}{{(z + i \sqrt{\frac{3}{2}})}^{8}}

= {\frac{(- 2 z + 2 i \sqrt{\frac{3}{2}}) (z + i \sqrt{\frac{3}{2}}) - 4 (- z^{2} + 2 i \sqrt{\frac{3}{2}} z + 3)}{{(z + i \sqrt{\frac{3}{2}})}^{5}}

2 R e s (φ, i \sqrt{\frac{3}{2}}) = {\frac{- 4 (\frac{3}{2} - 2 \frac{3}{2} + 3)}{{(2 i \sqrt{\frac{3}{2}})}^{5}}}

= \frac{- 6}{2^{5} i (\sqrt{\frac{3}{2}}) \frac{9}{4}} = \frac{- 6}{8 i \sqrt{\frac{3}{2}} . 9}

= \frac{- 2.3}{2.4 i . 3.3 \sqrt{\frac{3}{2}}} = - \frac{1}{4 i \frac{\sqrt{3}}{\sqrt{2}}} = \frac{- \sqrt{2}}{4 i \sqrt{3}}

R e s (φ, i \sqrt{\frac{3}{2}}) = \frac{- \sqrt{2}}{8 i \sqrt{3}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{- \sqrt{2}}{8 i \sqrt{3}} = \frac{- π \sqrt{2}}{4 \sqrt{3}} .

Commented by abdo.msup.com last updated on 04/Jun/18

e r r o r i n t h e f i n a l l i n e d

R e s (φ, i \sqrt{\frac{3}{2}}) = \frac{- \sqrt{2}}{12 i \sqrt{3}} a n d

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{- \sqrt{2}}{12 i \sqrt{3}}

= - \frac{\sqrt{2}}{6 \sqrt{3}} .