developp at integr serie ∫_(−∞) ^x (dt/(t^4 +t^2 +1))


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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
	$let ϕ(x) =∫_(−∞) ^x (dt/(t^4 +t^2 +1)) ⇒ϕ(x) =∫_(−∞) ^0 (dt/(t^4 +t^2 +1)) +∫_0 ^x (dt/(t^4 +t^2 +1)) ⇒ϕ^′ (x) =(1/(x^4 +x^2 +1)) poles of ϕ^′ ? x^4 +x^2 +1 =0 ⇒t^2 +t+1 =0 (x^2 =t) Δ =−3 ⇒t_1 =((−1+(√3))/2) =e^(i((2π)/3)) and t_2 =e^(−((i2π)/3)) ⇒ ϕ^′ (x) =(1/((x^2 −e^((i2π)/3) )(x^2 −e^(−((i2π)/3)) ))) =(1/((x−e^((iπ)/3) )(x+e^((iπ)/3) )(x−e^(−((iπ)/3)) )(x+e^(−((iπ)/3)) ))) =(a/(x−e^((iπ)/3) )) +(b/((x+e^((iπ)/3) ))) +(c/(x−e^(−((iπ)/3)) )) +(d/(x+e^(−((iπ)/3)) )) a =(1/(2e^((iπ)/3) (2i sin(((2π)/3))))) =(e^(−((iπ)/3)) /(4i ×((√3)/2))) =(e^(−((iπ)/3)) /(2i(√3))) b =(1/(−2e^((iπ)/3) (2isin(((2π)/3))))) =(e^(−((iπ)/3)) /(−4i×((√3)/2))) =−(e^(−((iπ)/3)) /(2i(√3))) c =(1/(2e^(−((iπ)/3)) (−2isin(((2π)/3))))) =−(e^((iπ)/3) /(4i×((√3)/2))) =−(e^((iπ)/3) /(2i(√3))) d =(1/(−2e^(−((iπ)/3)) (−2i sin(((2π)/3))))) =(e^((iπ)/3) /(2i(√3))) ⇒ ϕ^((n)) (x) =((a(−1)^(n−1) (n−1)!)/((x−e^((iπ)/3) )^n )) +((b(−1)^(n−1) (n−1)!)/((x+e^((iπ)/3) )^n )) +((c(−1)^(n−1) (n−1)!)/((x−e^(−((iπ)/3)) )^n )) +((d(−1)^(n−1) (n−1)!)/((x+e^(−((iπ)/3)) )^n )) ⇒ϕ^((n)) (0) = (−1)^(n−1) (n−1)! {a(−1)^n e^(−((inπ)/3)) +b e^(−((inπ)/3)) +c (−1)^n e^((inπ)/3) +d e^((inπ)/3) } ϕ(x) =Σ_(n=0) ^∞ ((ϕ^((n)) (0))/(n!)) x^n ....$
	$let φ (x) = \int_{- \infty}^{x} \frac{dt}{t^{4} + t^{2} + 1} \Rightarrow φ (x) = \int_{- \infty}^{0} \frac{dt}{t^{4} + t^{2} + 1} + \int_{0}^{x} \frac{dt}{t^{4} + t^{2} + 1}$ $\Rightarrow φ^{^{'}} (x) = \frac{1}{x^{4} + x^{2} + 1} poles of φ^{^{'}} ?$ $x^{4} + x^{2} + 1 = 0 \Rightarrow t^{2} + t + 1 = 0 (x^{2} = t)$ $Δ = - 3 \Rightarrow t_{1} = \frac{- 1 + \sqrt{3}}{2} = e^{i \frac{2 π}{3}} and t_{2} = e^{- \frac{i 2 π}{3}} \Rightarrow$ $φ^{^{'}} (x) = \frac{1}{(x^{2} - e^{\frac{i 2 π}{3}}) (x^{2} - e^{- \frac{i 2 π}{3}})} = \frac{1}{(x - e^{\frac{i π}{3}}) (x + e^{\frac{i π}{3}}) (x - e^{- \frac{i π}{3}}) (x + e^{- \frac{i π}{3}})}$ $= \frac{a}{x - e^{\frac{i π}{3}}} + \frac{b}{(x + e^{\frac{i π}{3}})} + \frac{c}{x - e^{- \frac{i π}{3}}} + \frac{d}{x + e^{- \frac{i π}{3}}}$ $a = \frac{1}{{2 e}^{\frac{i π}{3}} (2 i \sin (\frac{2 π}{3}))} = \frac{e^{- \frac{i π}{3}}}{4 i \times \frac{\sqrt{3}}{2}} = \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}$ $b = \frac{1}{- {2 e}^{\frac{i π}{3}} (2 isin (\frac{2 π}{3}))} = \frac{e^{- \frac{i π}{3}}}{- 4 i \times \frac{\sqrt{3}}{2}} = - \frac{e^{- \frac{i π}{3}}}{2 i \sqrt{3}}$ $c = \frac{1}{{2 e}^{- \frac{i π}{3}} (- 2 isin (\frac{2 π}{3}))} = - \frac{e^{\frac{i π}{3}}}{4 i \times \frac{\sqrt{3}}{2}} = - \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}}$ $d = \frac{1}{- {2 e}^{- \frac{i π}{3}} (- 2 i \sin (\frac{2 π}{3}))} = \frac{e^{\frac{i π}{3}}}{2 i \sqrt{3}} \Rightarrow$ $φ^{(n)} (x) = \frac{a {(- 1)}^{n - 1} (n - 1)!}{{(x - e^{\frac{i π}{3}})}^{n}} + \frac{b {(- 1)}^{n - 1} (n - 1)!}{{(x + e^{\frac{i π}{3}})}^{n}} + \frac{c {(- 1)}^{n - 1} (n - 1)!}{{(x - e^{- \frac{i π}{3}})}^{n}}$ $+ \frac{d {(- 1)}^{n - 1} (n - 1)!}{{(x + e^{- \frac{i π}{3}})}^{n}} \Rightarrow φ^{(n)} (0) =$ ${(- 1)}^{n - 1} (n - 1)! {a {(- 1)}^{n} e^{- \frac{in π}{3}} + b e^{- \frac{in π}{3}} + c {(- 1)}^{n} e^{\frac{in π}{3}} + d e^{\frac{in π}{3}}}$ $φ (x) = \sum_{n = 0}^{\infty} \frac{φ^{(n)} (0)}{n!} x^{n} . . . .$
	Commented by mathmax by abdo last updated on 21/May/20
	$anther way we have ϕ^′ (x) =(1/((x^2 −e^((i2π)/3) )(x^2 −e^(−((i2π)/3)) ))) =(1/(2isin((π/3))))((1/(x^2 −e^((i2π)/3) )) −(1/(x^2 −e^(−((i2π)/3)) )) ) =(1/(i(√3))){ (1/(x^2 −e^((i2π)/3) ))−(1/(x^2 −e^(−((i2π)/3)) ))} =(1/(i(√3)))( (e^(−((i2π)/3)) /(e^(−((i2π)/3)) x^2 −1)) −(e^((i2π)/3) /(e^((i2π)/3) −1))) =(1/(i(√3)))(− (e^(−((i2π)/3)) /(1−(e^(−((iπ)/3)) x)^2 )) +(e^((i2π)/3) /(1−(e^((iπ)/3) x)^2 ))) for ∣x∣<1 we get ϕ^′ (x) =(e^((i2π)/3) /(i(√3))) Σ_(n=0) ^∞ (e^((iπ)/3) x)^n −(e^(−((i2π)/3)) /(i(√3))) Σ_(n=0) ^∞ (e^(−((iπ)/3)) x)^n ϕ(x) =(e^((i2π)/3) /(i(√3))) Σ_(n=0) ^∞ e^((inπ)/3) (x^(n+1) /(n+1)) −(e^(−((i2π)/3)) /(i(√3))) Σ_(n=0) ^∞ e^(−((inπ)/3)) (x^(n+1) /(n+1)) +C$
	$anther way we have φ^{^{'}} (x) = \frac{1}{(x^{2} - e^{\frac{i 2 π}{3}}) (x^{2} - e^{- \frac{i 2 π}{3}})}$ $= \frac{1}{2 isin (\frac{π}{3})} (\frac{1}{x^{2} - e^{\frac{i 2 π}{3}}} - \frac{1}{x^{2} - e^{- \frac{i 2 π}{3}}}) = \frac{1}{i \sqrt{3}} {\frac{1}{x^{2} - e^{\frac{i 2 π}{3}}} - \frac{1}{x^{2} - e^{- \frac{i 2 π}{3}}}}$ $= \frac{1}{i \sqrt{3}} (\frac{e^{- \frac{i 2 π}{3}}}{e^{- \frac{i 2 π}{3}} x^{2} - 1} - \frac{e^{\frac{i 2 π}{3}}}{e^{\frac{i 2 π}{3}} - 1}) = \frac{1}{i \sqrt{3}} (- \frac{e^{- \frac{i 2 π}{3}}}{1 - {(e^{- \frac{i π}{3}} x)}^{2}} + \frac{e^{\frac{i 2 π}{3}}}{1 - {(e^{\frac{i π}{3}} x)}^{2}})$ $for ∣ x ∣< 1 we get φ^{^{'}} (x) = \frac{e^{\frac{i 2 π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} {(e^{\frac{i π}{3}} x)}^{n} - \frac{e^{- \frac{i 2 π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} {(e^{- \frac{i π}{3}} x)}^{n}$ $φ (x) = \frac{e^{\frac{i 2 π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} e^{\frac{in π}{3}} \frac{x^{n + 1}}{n + 1} - \frac{e^{- \frac{i 2 π}{3}}}{i \sqrt{3}} \sum_{n = 0}^{\infty} e^{- \frac{in π}{3}} \frac{x^{n + 1}}{n + 1} + C$
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