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

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

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}}