calculate-x-2-x-2-x-1-2-dx-

Question Number 117253 by mathmax by abdo last updated on 10/Oct/20

calculate \int_{- \infty}^{\infty} \frac{x^{2}}{{(x^{2} - x + 1)}^{2}} dx

Answered by mnjuly1970 last updated on 10/Oct/20

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

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

({\int_{- \infty}^{+ \infty} \frac{1}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x} = k_{1}) - ({[\frac{1}{2 (x^{2} - x + 1)}]}_{- \infty}^{\infty} = k_{2}) - \frac{1}{2} {\int_{- \infty}^{+ \infty} \frac{1}{{(x^{2} - x + 1)}^{2}} d x} = k_{3}

k_{1} = \frac{2 π}{\sqrt{3}}, k_{2} = 0

k_{3} = \int_{- \infty}^{+ \infty} \frac{1}{{(x^{2} - x + 1)}^{2}} d x

Λ (b) = \int_{- \infty}^{+ \infty} \frac{1}{x^{2} - x + b} d x = \int_{- \infty}^{+ \infty} \frac{1}{{(x - \frac{1}{2})}^{2} + {(\frac{\sqrt{4 b - 1}}{2})}^{2}}

= \frac{2 π}{\sqrt{4 b - 1}}

k_{3} = - Λ^{^{'}} (b) ∣_{b = 1} = - 2 π (\frac{- 1}{2}) (4) {(4 b - 1)}^{- \frac{3}{2}} ∣_{b = 1} = (4 π) \frac{1}{3 \sqrt{3}}

∴ \int_{- \infty}^{+ \infty} \frac{x^{2}}{{(x^{2} - x + 1)}^{2}} d x = \frac{2 π}{\sqrt{3}} - \frac{2 π}{3 \sqrt{3}} = \frac{4 π}{3 \sqrt{3}}

\dots . m . n . j u l y . 1970 \dots

Answered by Bird last updated on 10/Oct/20

l e t I = \int_{- \infty}^{+ \infty} \frac{x^{2}}{{(x^{2} - x + 1)}^{2}} d x a n d

φ (z) = \frac{z^{2}}{{(z^{2} - z + 1)}^{2}} p o l e s o f φ!

z^{2} - z + 1 = 0 \to Δ = - 3 \Rightarrow z_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}}

z_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}} \Rightarrow φ (z) = \frac{z^{2}}{{(z - e^{\frac{i π}{3}})}^{2} {(z - e^{- \frac{i π}{3}})}^{2}}

t h e p o l e s o f φ a r e e^{\frac{i π}{3}} a n d e^{- \frac{i π}{3}} (d o u b l e s)

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}{3}})

R e s (φ, e^{\frac{i π}{3}}) = {l i m}_{z \to e^{\frac{i π}{3}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{3}})}^{2} φ (z)}^{(1)}

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

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

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

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

= - 2 \times \frac{1}{{(2 i s i n (\frac{π}{3}))}^{3}}

= \frac{- 2}{- 8 i {(\frac{\sqrt{3}}{2})}^{3}} = \frac{1}{4 i (\frac{3 \sqrt{3}}{8})} = \frac{2}{3 i \sqrt{3}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π . \frac{2}{3 i \sqrt{3}} = \frac{4 π}{3 \sqrt{3}} \Rightarrow

I = \frac{4 π}{3 \sqrt{3}}

Answered by 1549442205PVT last updated on 11/Oct/20

\frac{x^{2}}{{(x^{2} - x + 1)}^{2}} \equiv \frac{ax + b}{(x^{2} - x + 1)} + \frac{{cx}^{2} + dx + e}{x^{4} - {2 x}^{3} + {3 x}^{2} - 2 x + 1}

\Leftrightarrow x^{2} \equiv {ax}^{5} + (c - 2 a) x^{4} + (3 a - 2 b - c + d) x^{3}

+ (3 b - 2 a + c - d + e) x^{2} +

+ (a - 2 b + d - e) x + b + e

\Leftrightarrow {\begin{cases} a = 0 \\ c - 2 a = 0 \\ 3 a - 2 b - c + d = 0 \\ - 2 a + 3 b + c - d + e = 1 \\ a - 2 b + d - e = 0 \\ b + e = 0 \end{cases}

\Leftrightarrow {\begin{cases} a = 0 \\ b = 2 / 3 = d \\ e = - 2 / 3 \\ c = 1 / 3 \end{cases}

Hence, \int_{- \infty}^{\infty} \frac{x^{2}}{{(x^{2} - x + 1)}^{2}} dx

= \frac{2}{3} \int_{- \infty}^{+ \infty} \frac{dx}{x^{2} - x + 1} + \frac{1}{3} \int_{- \infty}^{+ \infty} \frac{x^{2} + 2 x - 2}{{(x^{2} - x + 1)}^{2}} dx

Put F = \int \frac{x^{2} + 2 x - 2}{{(x^{2} - x + 1)}^{2}} dx . We will find F

in form : F (x) = \frac{{ax}^{2} + bx + c}{x^{2} - x + 1} + C

\Leftrightarrow \frac{x^{2} + 2 x - 2}{2 {(x^{2} - x + 1)}^{2}} \equiv F^{'} (x) = {(\frac{{ax}^{2} + bx + c}{x^{2} - x + 1})}^{'}

= \frac{(2 ax + b) (x^{2} - x + 1) - ({ax}^{2} + bx + c) (2 x - 1)}{{(x^{2} - x + 1)}^{2}}

= {{2 ax}^{3} + (b - 2 a) x^{2} + (2 a - b) x + b

- [{2 ax}^{3} + (2 b - a) x^{2} + (2 c - b) x - c]} / {(x^{2} - x + 1)}^{2}

= \frac{(- a - b) x^{2} + (2 a - 2 c) x + b + c}{{(x^{2} - x + 1)}^{2}}

\Leftrightarrow {\begin{cases} - a - b = 1 \\ 2 a - 2 c = 2 \\ b + c = - 2 \end{cases} \Leftrightarrow {\begin{cases} - 2 b - 2 c = 4 \\ b + c = - 2 \end{cases}

\Rightarrow b = - 2 - c

a = - b - 1 = 1 + c

\Rightarrow F (x) = \frac{(1 + c) x^{2} - (2 + c) x + c}{x^{2} - x + 1}

= \frac{- x - 1}{x^{2} - x + 1} + c + 1 (1) . We have

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

= \frac{2}{3} \times \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) (2) (since \int \frac{dx}{x^{2} + a^{2}} = \tan^{- 1} (\frac{x}{a}))

From (1) (2) and (c + 1) as a constant

We get : I = \int_{- \infty}^{\infty} \frac{x^{2}}{{(x^{2} - x + 1)}^{2}} dx

= {[\frac{4}{3 \sqrt{3}} \tan^{- 1} (\frac{2 x - 1}{\sqrt{3}}) - \frac{1}{3} . \frac{x + 1}{x^{2} - x + 1}]}_{- \infty}^{+ \infty}

= \frac{4 \sqrt{3}}{9} \times \frac{π}{2} \times 2 = \frac{4 π \sqrt{3}}{9}

Commented by 1549442205PVT last updated on 11/Oct/20

Thank Prof . You are welcome .

Commented by mathmax by abdo last updated on 11/Oct/20

thank you sir .