calculate-dx-x-2-3x-7-2-

Question Number 76359 by mathmax by abdo last updated on 26/Dec/19

c a l c u l a t e \int_{- \infty}^{\infty} \frac{d x}{{(x^{2} - 3 x + 7)}^{2}}

Commented by mathmax by abdo last updated on 28/Dec/19

l e t I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - 3 x + 7)}^{2}} a n d φ (z) = \frac{1}{{(z^{2} - 3 z + 7)}^{2}}] p o l e s o f φ ?

z^{2} - 3 z + 7 = 0 \to Δ = 9 - 4.7 = 9 - 28 = - 19

z_{1} = \frac{3 + i \sqrt{19}}{2} a n d z_{2} = \frac{3 - i \sqrt{19}}{2}

φ (z) = \frac{1}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} 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, z_{1})

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

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

= - 2 {(z_{1} - z_{2})}^{- 3} = - 2 {(i \sqrt{19})}^{- 3} = \frac{- 2}{{(i \sqrt{19})}^{3}} = \frac{- 2}{- i (19) \sqrt{19}} = \frac{2}{19 i \sqrt{19}}

\Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \times \frac{2}{19 i \sqrt{19}} = \frac{4 π}{19 \sqrt{19}} = I