calculate-I-1-x-2-1-x-4-x-2-1-dx-

Question Number 86258 by lémùst last updated on 27/Mar/20

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

Commented by mathmax by abdo last updated on 27/Mar/20

I = \int_{1}^{+ \infty} \frac{x^{2} - 1}{x^{4} - x^{2} + 1} d x (c o m p l e x m e t h o d)

x^{4} - x^{2} + 1 = 0 \to t^{2} - t + 1 = 0 (t = x^{2})

Δ = 1 - 4 = - 3 \Rightarrow t_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{6}} a n d t_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{6}}

\Rightarrow \frac{x^{2} - 1}{x^{4} - x^{2} + 1} = \frac{x^{2} - 1}{(x^{2} - e^{\frac{i π}{6}}) (x^{2} - e^{- \frac{i π}{6}})} = \frac{x^{2} - 1}{2 i s i n (\frac{π}{6})} {\frac{1}{x^{2} - e^{\frac{i π}{6}}} - \frac{1}{x^{2} - e^{- \frac{i π}{6}}}}

= - i (\frac{x^{2} - 1}{x^{2} - e^{\frac{i π}{6}}} - \frac{x^{2} - 1}{x^{2} - e^{- \frac{i π}{6}}}) = - i (\frac{x^{2} - e^{\frac{i π}{6}} + e^{\frac{i π}{6}} - 1}{x^{2} - e^{\frac{i π}{6}}} - \frac{x^{2} - e^{- \frac{i π}{6}} + e^{- \frac{i π}{6}} - 1}{x^{2} - e^{- \frac{i π}{6}}})

= - i {\frac{e^{\frac{i π}{6}} - 1}{x^{2} - e^{\frac{i π}{6}}} - \frac{e^{- \frac{i π}{6}} - 1}{x^{2} - e^{- \frac{i π}{6}}}} \Rightarrow I = i (e^{- \frac{i π}{6}} - 1) \int_{1}^{+ \infty} \frac{d x}{x^{2} - e^{- \frac{i π}{6}}}

- i (e^{\frac{i π}{6}} - 1) \int_{1}^{+ \infty} \frac{d x}{x^{2} - e^{\frac{i π}{6}}}

\int_{1}^{+ \infty} \frac{d x}{x^{2} - e^{\frac{i π}{6}}} = \int_{1}^{+ \infty} \frac{d x}{(x - e^{\frac{i π}{12}}) (x + e^{\frac{i π}{12}})}

= \frac{1}{2} e^{- \frac{i π}{12}} \int_{1}^{+ \infty} (\frac{1}{x - e^{\frac{i π}{12}}} - \frac{1}{x + e^{\frac{i π}{12}}}) d x

= \frac{1}{2} e^{- \frac{i π}{12}} {[l n (\frac{x - e^{\frac{i π}{12}}}{x + e^{\frac{i π}{12}}})]}_{1}^{+ \infty} = \frac{1}{2} e^{- \frac{i π}{12}} {- l n (\frac{1 - e^{\frac{i π}{12}}}{1 + e^{\frac{i π}{12}}})}

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

Commented by lémùst last updated on 28/Mar/20

s i r h o w d o y o u c a l c u l a t e t h e l o g a r i t h o f a

c o m p l e x ?

Commented by mathmax by abdo last updated on 31/Mar/20

l n (x + i y) = l n (\sqrt{x^{2} + y^{2}} e^{i a r c t a n (\frac{y}{x})})

= \frac{1}{2} l n (x^{2} + y^{2}) + i a r c t a n (\frac{y}{x}) (p r i n c i p l e d e t e r m i n a t i o n o f l n)

Answered by TANMAY PANACEA. last updated on 28/Mar/20

\int \frac{1 - \frac{1}{x^{2}}}{x^{2} - 1 + \frac{1}{x^{2}}} d x

\int \frac{d (x + \frac{1}{x})}{{(x + \frac{1}{x})}^{2} - 3} = \frac{1}{2 \sqrt{3}} l n (\frac{x + \frac{1}{x} - \sqrt{3}}{x + \frac{1}{x} + \sqrt{3}})

u s e i n g f o r m u l a \int \frac{d p}{p^{2} - a^{2}}

Commented by lémùst last updated on 28/Mar/20

t h a n k s s i r

Commented by TANMAY PANACEA. last updated on 28/Mar/20

m o s t w e l c o m e \dots