calculate-0-dx-x-4-x-2-1-2-

Question Number 83253 by mathmax by abdo last updated on 29/Feb/20

c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(x^{4} - x^{2} + 1)}^{2}}

Commented by abdomathmax last updated on 01/Mar/20

s o r r y t h e Q i s c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{4} - x^{2} + 1)}^{2}}

Commented by mathmax by abdo last updated on 01/Mar/20

l e t I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{4} - x^{2} + 1)}^{2}} l e t φ (z) = \frac{1}{{(z^{4} - z^{2} + 1)}^{2}} p o l e s o f φ ?

z^{4} - z^{2} + 1 = 0 \Rightarrow t^{2} - t + 1 = 0 (t = z^{2})

Δ = 1 - 4 = - 3 \Rightarrow t_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} a n d t_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}}

t^{2} - t + 1 = (t - e^{\frac{i π}{3}}) (t - e^{- \frac{i π}{3}}) \Rightarrow z^{4} - z^{2} + 1 = (z^{2} - e^{\frac{i π}{3}}) (z^{2} - e^{\frac{- i π}{3}})

= (z - e^{\frac{i π}{6}}) (z + e^{\frac{i π}{6}}) (z - e^{- \frac{i π}{6}}) (z + e^{\frac{- i π}{6}}) \Rightarrow

φ (z) = \frac{1}{{(z - e^{\frac{i π}{6}})}^{2} {(z + e^{\frac{i π}{6}})}^{2} {(z - e^{- \frac{i π}{6}})}^{2} {(z + e^{- \frac{i π}{6}})}^{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, e^{\frac{i π}{6}}) + R e s (φ, - e^{- \frac{i π}{6}})}

R e s (φ, e^{\frac{i π}{6}}) = {l i m}_{z \to e^{\frac{i π}{6}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{6}})}^{2} φ (z)}^{(1)}

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

= {l i m}_{z \to e^{\frac{i π}{6}}} - \frac{{(z^{2} - e^{- \frac{i π}{3}})}^{2} + (z + e^{\frac{i π}{6}}) \times (4 z) (z^{2} - e^{- \frac{i π}{3}})}{{(z + e^{\frac{i π}{6}})}^{2} {(z^{2} - e^{- \frac{i π}{3}})}^{4}}

= - {l i m}_{z \to e^{\frac{i π}{6}}} \frac{(z^{2} - e^{- \frac{i π}{3}}) + 4 z (z + e^{\frac{i π}{6}})}{{(z + e^{\frac{i π}{6}})}^{2} {(z^{2} - e^{- \frac{i π}{3}})}^{3}}

= - \frac{e^{\frac{i π}{3}} - e^{- \frac{i π}{3}} + 4 e^{\frac{i π}{6}} \times 2 e^{\frac{i π}{6}}}{{(2 e^{\frac{i π}{6}})}^{2} {(e^{\frac{i π}{3}} - e^{- \frac{i π}{3}})}^{3}} = - \frac{2 i s i n (\frac{π}{3}) + 8 e^{\frac{i π}{3}}}{2 e^{\frac{i π}{3}} {(2 i)}^{3} {s i n}^{3} (\frac{π}{3})}

= - \frac{i s i n (\frac{π}{3}) + 4 e^{\frac{i π}{3}}}{- 8 i e^{\frac{i π}{3}} {s i n}^{3} (\frac{π}{3})} = e^{- \frac{i π}{3}} \times \frac{s i n (\frac{π}{3}) - 4 {i e}^{\frac{i π}{3}}}{{s i n}^{3} (\frac{π}{3})}

= \frac{s i n (\frac{π}{3}) e^{- \frac{i π}{3}} - 4 i}{8 {(\frac{\sqrt{3}}{2})}^{3}} = \frac{\frac{\sqrt{3}}{2} e^{- \frac{i π}{3}} - 4 i}{3 \sqrt{3}} \dots . b e c o n t i n u e d \dots

Commented by mathmax by abdo last updated on 01/Mar/20

p a r e a m e t r i c m e t h o d l e t φ (a) = \int_{0}^{\infty} \frac{d x}{x^{4} - x^{2} + a} w i t h a > \frac{1}{4}

w e h a v e φ^{^{'}} (a) = - \int_{0}^{\infty} \frac{d x}{{(x^{4} - x^{2} + a)}^{2}} \Rightarrow \int_{0}^{\infty} \frac{d x}{{(x^{4} - x^{2} + a)}^{2}} = - φ^{^{'}} (a)

w e h a v e 2 φ (a) = \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} - x^{2} + a} l e t W (z) = \frac{1}{z^{4} - z^{2} + a}

p o l e s o f W ?

z^{4} - z^{2} + a = 0 \Rightarrow t^{2} - t + a = 0 (t = z^{2})

Δ = 1 - 4 a < 0 \Rightarrow Δ = {(i \sqrt{4 a - 1})}^{2} \Rightarrow z_{1} = \frac{1 + i \sqrt{4 a - 1}}{2}

z_{2} = \frac{1 - i \sqrt{4 a - 1}}{2} \Rightarrow W (z) = \frac{1}{(z^{2} - z_{1}) (z^{2} - z_{2})} = \frac{1}{(z - \sqrt{z_{1}}) (z + \sqrt{z_{1}}) (z - \sqrt{z_{2}}) (z + \sqrt{z_{2}})}

\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, \sqrt{z_{1}}) + R e s (W, - \sqrt{z_{2}})}

R e s (W, \sqrt{z_{1}}) = \frac{1}{2 \sqrt{z_{1}} (z_{1}^{2} - z_{2})}

R e s (W, - \sqrt{z_{2}}) = \frac{1}{- 2 \sqrt{z_{2}} (z_{2}^{2} - z_{1})} = - \frac{1}{2 \sqrt{z_{2}} (z_{2}^{2} - z_{1})} \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = i π {\frac{1}{\sqrt{z_{1}} (z_{1}^{2} - z_{2})} - \frac{1}{\sqrt{z_{2}} (z_{2}^{2} - z_{1})}}

∣ z_{1} ∣= \frac{1}{2} \sqrt{1 + 4 a - 1} = \sqrt{a} \Rightarrow z_{1} = \sqrt{a} e^{i a r c t a n \sqrt{4 a - 1}}

z_{2} = \sqrt{a} e^{- i a r c t a n (\sqrt{4 a - 1})} \Rightarrow \sqrt{z_{1}} =^{4} \sqrt{a} e^{\frac{i}{2} a r c t a n (\sqrt{4 a - 1})}

z_{1}^{2} - z_{2} = a e^{2 i a r c t a n (\sqrt{4 a - 1})} - \sqrt{a} e^{- i a r c t a n (\sqrt{4 a - 1})} \Rightarrow

\int_{- \infty}^{+ \infty} W (z) d z = i π {\frac{1}{(^{4} \sqrt{a}) e^{i a r c t a n (\sqrt{4 a - 1})} (a e^{2 i a r c t a n (\sqrt{4 a - 1})} - \sqrt{a} e^{- i a r c t a n (\sqrt{4 a - 1})}} - \frac{1}{(^{4} \sqrt{a}) e^{- i a r c t a n (\sqrt{4 a - 1})} ({a e}^{- 2 i a r c t a n (\sqrt{4 a - 1})} - \sqrt{a} e^{i a r c t a n (\sqrt{4 a - 1})})}}

\dots b e c o n t i n u e d \dots