developp-at-integr-serie-x-dt-t-4-t-2-1-

Question Number 94333 by mathmax by abdo last updated on 18/May/20

d e v e l o p p a t i n t e g r s e r i e \int_{- \infty}^{x} \frac{d t}{t^{4} + t^{2} + 1}

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

let φ (x) = \int_{- \infty}^{x} \frac{dt}{t^{4} + t^{2} + 1} \Rightarrow φ (x) = \int_{- \infty}^{0} \frac{dt}{t^{4} + t^{2} + 1} + \int_{0}^{x} \frac{dt}{t^{4} + t^{2} + 1}

\Rightarrow φ^{^{'}} (x) = \frac{1}{x^{4} + x^{2} + 1} poles of φ^{^{'}} ?

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

Δ = - 3 \Rightarrow t_{1} = \frac{- 1 + \sqrt{3}}{2} = e^{i \frac{2 π}{3}} and t_{2} = e^{- \frac{i 2 π}{3}} \Rightarrow

φ^{^{'}} (x) = \frac{1}{(x^{2} - e^{\frac{i 2 π}{3}}) (x^{2} - e^{- \frac{i 2 π}{3}})} = \frac{1}{(x - e^{\frac{i π}{3}}) (x + e^{\frac{i π}{3}}) (x - e^{- \frac{i π}{3}}) (x + e^{- \frac{i π}{3}})}

= \frac{a}{x - e^{\frac{i π}{3}}} + \frac{b}{(x + e^{\frac{i π}{3}})} + \frac{c}{x - e^{- \frac{i π}{3}}} + \frac{d}{x + e^{- \frac{i π}{3}}}

a = \frac{1}{{2 e}^{\frac{i π}{3}} (2 i \sin (\frac{2 π}{3}))} = \frac{e^{- \frac{i π}{3}}}{4 i \times \frac{\sqrt{3}}{2}} = \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}

b = \frac{1}{- {2 e}^{\frac{i π}{3}} (2 isin (\frac{2 π}{3}))} = \frac{e^{- \frac{i π}{3}}}{- 4 i \times \frac{\sqrt{3}}{2}} = - \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}

c = \frac{1}{{2 e}^{- \frac{i π}{3}} (- 2 isin (\frac{2 π}{3}))} = - \frac{e^{\frac{i π}{3}}}{4 i \times \frac{\sqrt{3}}{2}} = - \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}}

d = \frac{1}{- {2 e}^{- \frac{i π}{3}} (- 2 i \sin (\frac{2 π}{3}))} = \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}} \Rightarrow

φ^{(n)} (x) = \frac{a {(- 1)}^{n - 1} (n - 1)!}{{(x - e^{\frac{i π}{3}})}^{n}} + \frac{b {(- 1)}^{n - 1} (n - 1)!}{{(x + e^{\frac{i π}{3}})}^{n}} + \frac{c {(- 1)}^{n - 1} (n - 1)!}{{(x - e^{- \frac{i π}{3}})}^{n}}

+ \frac{d {(- 1)}^{n - 1} (n - 1)!}{{(x + e^{- \frac{i π}{3}})}^{n}} \Rightarrow φ^{(n)} (0) =

{(- 1)}^{n - 1} (n - 1)! {a {(- 1)}^{n} e^{- \frac{in π}{3}} + b e^{- \frac{in π}{3}} + c {(- 1)}^{n} e^{\frac{in π}{3}} + d e^{\frac{in π}{3}}}

φ (x) = \sum_{n = 0}^{\infty} \frac{φ^{(n)} (0)}{n!} x^{n} \dots .

Commented by mathmax by abdo last updated on 21/May/20

anther way we have φ^{^{'}} (x) = \frac{1}{(x^{2} - e^{\frac{i 2 π}{3}}) (x^{2} - e^{- \frac{i 2 π}{3}})}

= \frac{1}{2 isin (\frac{π}{3})} (\frac{1}{x^{2} - e^{\frac{i 2 π}{3}}} - \frac{1}{x^{2} - e^{- \frac{i 2 π}{3}}}) = \frac{1}{i \sqrt{3}} {\frac{1}{x^{2} - e^{\frac{i 2 π}{3}}} - \frac{1}{x^{2} - e^{- \frac{i 2 π}{3}}}}

= \frac{1}{i \sqrt{3}} (\frac{e^{- \frac{i 2 π}{3}}}{e^{- \frac{i 2 π}{3}} x^{2} - 1} - \frac{e^{\frac{i 2 π}{3}}}{e^{\frac{i 2 π}{3}} - 1}) = \frac{1}{i \sqrt{3}} (- \frac{e^{- \frac{i 2 π}{3}}}{1 - {(e^{- \frac{i π}{3}} x)}^{2}} + \frac{e^{\frac{i 2 π}{3}}}{1 - {(e^{\frac{i π}{3}} x)}^{2}})

for ∣ x ∣< 1 we get φ^{^{'}} (x) = \frac{e^{\frac{i 2 π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} {(e^{\frac{i π}{3}} x)}^{n} - \frac{e^{- \frac{i 2 π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} {(e^{- \frac{i π}{3}} x)}^{n}

φ (x) = \frac{e^{\frac{i 2 π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} e^{\frac{in π}{3}} \frac{x^{n + 1}}{n + 1} - \frac{e^{- \frac{i 2 π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} e^{- \frac{in π}{3}} \frac{x^{n + 1}}{n + 1} + C