calculate-dx-1-x-x-2-3-

Question Number 35055 by math khazana by abdo last updated on 14/May/18

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(1 + x + x^{2})}^{3}}

Commented by math khazana by abdo last updated on 15/May/18

l e t p u t I = \int_{- \infty}^{+ \infty} \frac{d x}{{(1 + x + x^{2})}^{3}}

I = \int_{- \infty}^{+ \infty} \frac{d x}{{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}^{3}} ? c h a n g e m e n t

x + \frac{1}{2} = \frac{\sqrt{3}}{2} t a n t g i b e

I = \int_{- \frac{π}{3}}^{\frac{π}{2}} \frac{1}{{(\frac{3}{4})}^{3} {(1 + {t a n}^{2} t)}^{3}} \frac{\sqrt{3}}{2} (1 + {t a n}^{2} t) d t

= \frac{4^{3}}{3^{3}} \frac{\sqrt{3}}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d t}{{(1 + {t a n}^{2} t)}^{2}}

= \frac{4^{3}}{3^{3}} \sqrt{3} \int_{0}^{\frac{π}{2}} {c o s}^{4} t d t

= \frac{4^{3}}{3^{3}} \sqrt{3} \int_{0}^{\frac{π}{2}} \frac{{(1 + c o s (2 t))}^{2}}{4} d t

= \frac{16}{27} \sqrt{3} \int_{0}^{\frac{π}{2}} (1 + 2 c o s (2 t) + \frac{1 + c o s (4 t)}{2}) d t

= \frac{8 π \sqrt{3}}{27} + \frac{4 π \sqrt{3}}{27} = \frac{12 π \sqrt{3}}{3.9} = \frac{4 π \sqrt{3}}{9}

I = \frac{4 π \sqrt{3}}{9} .

Commented by abdo imad last updated on 17/May/18

l e t i n t r o d u c e t h e c o m p l e x f u n c t i o n φ (z) = \frac{1}{{(1 + z + z^{2})}^{3}}

t h e [p o l e s o f φ a r e j = e^{i \frac{2 π}{3}} a n d \bar{j} = e^{- i \frac{2 π}{3}} (t r i p l e p o l e s

φ (z) = \frac{1}{{(z - j)}^{3} {(z - \bar{j})}^{3}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, j) b u t

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

= {l i m}_{z \to j} \frac{1}{2} {{(z - \bar{j})}^{- 3}}^{(2)} b u t w e h a v e

{{(z - \bar{j})}^{- 3}}^{(1)} = - 3 {(z - \bar{j})}^{- 4}

{(z - \bar{j})}^{(2)} = 12 {(z - \bar{j})}^{- 5}

R e s (φ, j) = {l i m}_{z \to j} 6 {(z - \bar{j})}^{- 5} = 6 {(j - \bar{j})}^{- 5}

= 6 {(2 i \frac{\sqrt{3}}{2})}^{- 5} = \frac{6}{{(i \sqrt{3})}^{5}} = \frac{6}{i 9 \sqrt{3}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{6}{i 9 \sqrt{3}} = \frac{2 π . 3.2}{3.3 \sqrt{3}} = \frac{4 π}{3 \sqrt{3}} \Rightarrow

★ \int_{- \infty}^{+ \infty} \frac{d x}{{(1 + x + x^{2})}^{3}} = \frac{4 π \sqrt{3}}{9} ★

Answered by MJS last updated on 15/May/18

\int \frac{d x}{{(x^{2} + x + 1)}^{3}} = \int \frac{d x}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{3}} =

= 64 \int \frac{d x}{{({(2 x + 1)}^{2} + + 3)}^{3}} =

[u = 2 x + 1 \to d x = \frac{d u}{2}]

= 32 \int \frac{d u}{{(u^{2} + 3)}^{3}}

[\int \frac{d u}{{({a u}^{2} + b)}^{n}} = \frac{u}{2 b (n - 1) {({a u}^{2} + b)}^{n - 1}} + \frac{2 n - 3}{2 b (n - 1)} \int \frac{d u}{{({a u}^{2} + b)}^{n - 1}}]

a = 1; b = 3; n = 3

= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} \int \frac{d u}{{(u^{2} + 3)}^{2}}) =

a = 1; b = 3; n = 2

= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} (\frac{u}{6 (u^{2} + 3)} + \frac{1}{6} \int \frac{d u}{u^{2} + 3})) =

[v = \frac{u \sqrt{3}}{3} \to d u = \sqrt{3} d v]

= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} (\frac{u}{6 (u^{2} + 3)} + \frac{\sqrt{3}}{18} \int \frac{d v}{v^{2} + 1})) =

= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} (\frac{u}{6 (u^{2} + 3)} + \frac{\sqrt{3}}{18} \arctan (v))) =

= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} (\frac{u}{6 (u^{2} + 3)} + \frac{\sqrt{3}}{18} \arctan (\frac{u \sqrt{3}}{3}))) =

= \frac{8 u}{3 {(u^{2} + 3)}^{2}} + \frac{4 u}{3 (u^{2} + 3)} + \frac{4 \sqrt{3}}{9} \arctan (\frac{u \sqrt{3}}{3}) =

= \frac{4 u (u^{2} + 5)}{3 {(u^{2} + 3)}^{2}} + \frac{4 \sqrt{3}}{9} \arctan (\frac{u \sqrt{3}}{3}) =

= \frac{4 (2 x + 1) ({(2 x + 1)}^{2} + 5)}{3 {({(2 x + 1)}^{2} + 3)}^{2}} + \frac{4 \sqrt{3}}{9} \arctan (\frac{\sqrt{3}}{3} (2 x + 1)) =

= \frac{(2 x + 1) (2 x^{2} + 2 x + 3)}{6 {(x^{2} + x + 1)}^{2}} + \frac{4 \sqrt{3}}{9} \arctan (\frac{\sqrt{3}}{3} (2 x + 1)) + C

\underset{- \infty}{\int^{\infty}} \frac{d x}{{(x^{2} + x + 1)}^{3}} = \frac{4 \sqrt{3}}{9} π

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