calculate ∫_(−∞) ^(+∞) (dx/((x^2 −x+1)^3 ))


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 66693 by mathmax by abdo last updated on 18/Aug/19

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - x + 1)}^{3}}$
Commented by mathmax by abdo last updated on 19/Aug/19
$let A =∫_(−∞) ^(+∞) (dx/((x^2 −x+1)^3 )) wehave x^2 −x+1 =(x−(1/2))^2 +(3/4) we do the changement x−(1/2) =((√3)/2)t ⇒ A =((4/3))^3 ∫_(−∞) ^(+∞) (1/((t^2 +1)^3 ))((√3)/2)dt =((√3)/2) ((64)/(27)) =((32(√3))/(27)) ∫_(−∞) ^(+∞) (dt/((t^2 +1)^3 )) let W(z) =(1/((z^2 +1)^3 )) ⇒ W(z) =(1/((z−i)^3 (z+i)^3 )) so the poles of W are +^− i (triples) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) Res(W,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 W(z)}^((2)) =lim_(z→i) (1/2){(z+i)^(−3) }^((2)) =(1/2)lim_(z→i) {−3(z+i)^(−4) }^((1)) =−(3/2)lim_(z→i) {−4(z+i)^(−5) } =6(2i)^(−5) =(6/(2^5 i^5 )) =(3/(16 i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ (3/(16i)) =((3π)/8) ⇒ A =((32(√3))/(27))×((3π)/8) =((4π(√3))/9)$
$l e t A = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} - x + 1)}^{3}} w e h a v e x^{2} - x + 1 = {(x - \frac{1}{2})}^{2} + \frac{3}{4}$ $w e d o t h e c h a n g e m e n t x - \frac{1}{2} = \frac{\sqrt{3}}{2} t \Rightarrow A = {(\frac{4}{3})}^{3} \int_{- \infty}^{+ \infty} \frac{1}{{(t^{2} + 1)}^{3}} \frac{\sqrt{3}}{2} d t$ $= \frac{\sqrt{3}}{2} \frac{64}{27} = \frac{32 \sqrt{3}}{27} \int_{- \infty}^{+ \infty} \frac{d t}{{(t^{2} + 1)}^{3}} l e t W (z) = \frac{1}{{(z^{2} + 1)}^{3}} \Rightarrow$ $W (z) = \frac{1}{{(z - i)}^{3} {(z + i)}^{3}} s o t h e p o l e s o f W a r e \bar{+} i (t r i p l e s)$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i)$ $R e s (W, i) = {l i m}_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} W (z)}^{(2)}$ $= {l i m}_{z \to i} \frac{1}{2} {{(z + i)}^{- 3}}^{(2)} = \frac{1}{2} {l i m}_{z \to i} {- 3 {(z + i)}^{- 4}}^{(1)}$ $= - \frac{3}{2} {l i m}_{z \to i} {- 4 {(z + i)}^{- 5}} = 6 {(2 i)}^{- 5} = \frac{6}{2^{5} i^{5}} = \frac{3}{16 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π \frac{3}{16 i} = \frac{3 π}{8} \Rightarrow A = \frac{32 \sqrt{3}}{27} \times \frac{3 π}{8} = \frac{4 π \sqrt{3}}{9}$
	Answered by mind is power last updated on 18/Aug/19

	$l e t f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{(x^{2} - x + a)} w i t h e a > \frac{1}{4}$ $f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{({(x - \frac{1}{2})}^{2} + \frac{4 a - 1}{4})}$ $f (a) = \int_{- \infty}^{+ \infty} \frac{d x}{({(\frac{2 x - 1}{\sqrt{(4 a - 1)}})}^{2} + 1)} . \frac{4}{4 a - 1}$ $f (a) = \frac{2}{\sqrt{4 a - 1}} \int_{- \infty}^{+ \infty} \frac{d \frac{2 x - 1}{\sqrt{4 a - 1}}}{({(\frac{2 x - 1}{\sqrt{4 a - 1})})}^{2} + 1)}$ $= f (a) = \frac{2}{\sqrt{4 a - 1}} {[a r c t g (\frac{2 x - 1}{\sqrt{4 a - 1}})]}_{- \infty}^{+ \infty} = \frac{2 π}{\sqrt{4 a - 1}}$ $f^{'} (a) = \frac{d}{d a} \int_{- \infty}^{+ \infty} \frac{1}{x^{2} - x + a} d x = \int_{- \infty}^{+ \infty} - \frac{d x}{{(x^{2} - x + a)}^{2}}$ $f^{″} (a) = \int_{- \infty}^{+ \infty} \frac{2 d x}{{(a + x + x^{2})}^{3}}$ $\int_{- \infty}^{+ \infty} \frac{2 d x}{{(1 + x + x^{2})}^{3}} = f^{″} (1)$ $f^{'} (a) = {(\frac{2 π}{\sqrt{4 a - 1}})}^{'} = - 4 π {(4 a - 1)}^{\frac{- 3}{2}}$ $f^{″} (a) = 24 π {(4 a - 1)}^{- \frac{5}{2}} . . > f^{″} (1) = 24 π {(3)}^{- \frac{5}{2}}$ $==> 2 \int_{- \infty}^{+ \infty} \frac{d x}{{(1 + x + x^{2})}^{3}} = \frac{24 π}{9 \sqrt{3}} ==> \int_{- \infty}^{+ \infty} \frac{d x}{{(1 + x + x^{2})}^{3}} = \frac{4 π}{3 \sqrt{3}}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum