1-calculste-A-dx-x-2-ix-1-2-2-extract-Re-A-and-Im-A-and-determines-its-values-

Question Number 112449 by mathmax by abdo last updated on 07/Sep/20

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

2) extract Re (A) and Im (A) and determines its values

Commented by MJS_new last updated on 08/Sep/20

I get A = \frac{4 π \sqrt{5}}{25} can this be true ?

Commented by MJS_new last updated on 08/Sep/20

Here's a better place to answer you. I'm well and I hope you're well, too. Nice to be in here again.

Commented by abdomsup last updated on 08/Sep/20

y o u a r e w e l c o m e s i r w i s h y o u

h a p p y r e t u r n . . i h a v e n t t h e v a l u e

o f t h i s i n t e g r a l b u t w a i t m y a n s w e r

s o o n \dots .

Answered by mathmax by abdo last updated on 08/Sep/20

1) A = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - ix + 1)}^{2}} let φ (z) = \frac{1}{{(z^{2} - iz + 1)}^{2}} poles of φ ?

z^{2} - iz + 1 = 0 \to Δ = {(- i)}^{2} - 4 = - 5 \Rightarrow z_{1} = \frac{i + i \sqrt{5}}{2} and z_{2} = \frac{i - i \sqrt{5}}{2}

\Rightarrow φ (z) = \frac{1}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}} residus tbeorem give

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, z_{1}) and

Res (φ, z_{1}) = \lim_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}

= \lim_{z \to z_{1}} {\frac{1}{{(z - z_{2})}^{2}}}^{(1)} = - \lim_{z \to z_{1}} \frac{2 (z - z_{2})}{{(z - z_{2})}^{4}}

= \lim_{z \to z_{1}} \frac{- 2}{{(z - z_{2})}^{3}} = \frac{- 2}{{(z_{1} - z_{2})}^{3}} = \frac{- 2}{{(i \sqrt{5})}^{3}} = \frac{- 2}{- i 5 \sqrt{5}} = \frac{2}{5 i \sqrt{5}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π \times \frac{2}{5 i \sqrt{5}} = \frac{4 π}{5 \sqrt{5}} \Rightarrow ★ A = \frac{4 π}{5 \sqrt{5}} ★

Answered by MJS_new last updated on 08/Sep/20

\frac{1}{{(x^{2} - i x + 1)}^{2}} =

= \frac{x^{4} + x^{2} + 1}{{(x^{4} + 3 x^{2} + 1)}^{2}} - \frac{2 x (x^{2} + 1)}{{(x^{4} + 3 x^{2} + 1)}^{2}} i

\Rightarrow the imaginary part is uneven

\Rightarrow \underset{- a}{\int^{+ a}} \frac{d x}{{(x^{2} - i x + 1)}^{2}} = \underset{- a}{\int^{+ a}} \frac{x^{4} + x^{2} + 1}{{(x^{4} + 3 x^{2} + 1)}^{2}} d x

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

[Ostrogradski]

= \frac{x (2 x^{2} + 3)}{5 (x^{4} + 3 x^{2} + 1)} + \frac{2}{5} \int \frac{x^{2} + 1}{x^{4} + 3 x^{2} + 1} d x

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

= \frac{5 + \sqrt{5}}{10} \int \frac{d x}{x^{2} + \frac{3 + \sqrt{5}}{2}} + \frac{5 - \sqrt{5}}{10} \int \frac{d x}{x^{2} + \frac{3 - \sqrt{5}}{2}} =

= \frac{\sqrt{5}}{5} (- \arctan \frac{(1 - \sqrt{5}) x}{2} + \arctan \frac{(1 + \sqrt{5}) x}{2})

\Rightarrow

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

= \frac{x (2 x^{2} + 3)}{5 (x^{4} + 3 x^{2} + 1)} + \frac{2 \sqrt{5}}{25} (\arctan \frac{(1 + \sqrt{5}) x}{2} - \arctan \frac{(1 - \sqrt{5}) x}{2}) + C

\Rightarrow

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

= \frac{4 π \sqrt{5}}{25}

Commented by mathmax by abdo last updated on 08/Sep/20

thank you sir M^{j} S

Answered by mathmax by abdo last updated on 08/Sep/20

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

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

= \int_{- \infty}^{+ \infty} \frac{x^{4} + {2 x}^{2} + 1 + {2 ix}^{3} + 2 ix - x^{2}}{{(x^{4} + {2 x}^{2} + 1 + x^{2})}^{2}} dx

= \int_{- \infty}^{+ \infty} \frac{x^{4} + x^{2} + 1 + i ({2 x}^{3} + 2 x)}{{(x^{4} + {3 x}^{2} + 1)}^{2}} dx \Rightarrow

Re (A) = \int_{- \infty}^{+ \infty} \frac{x^{4} + x^{2} + 1}{{(x^{4} + {3 x}^{2} + 1)}^{2}} dx and Im (A) = \int_{- \infty}^{+ \infty} \frac{{2 x}^{3} + 2 x}{{(x^{4} + {3 x}^{2} + 1)}^{2}} dx

we conclude that Res (A) = \frac{4 π}{5 \sqrt{5}} and Im (A) = 0