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

Question Number 162016 by mathmax by abdo last updated on 25/Dec/21

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

Commented by MJS_new last updated on 25/Dec/21

I can solve the indefinite integral but {it}^{'} s a

long hard way \dots

Commented by mathmax by abdo last updated on 25/Dec/21

use residus theorem sir

Answered by Ar Brandon last updated on 24/Mar/22

Υ = \int_{- \infty}^{+ \infty} \frac{\cos 3 x}{{(x^{2} + x + 1)}^{2}} d x = R e \int_{- \infty}^{+ \infty} \frac{e^{3 i x}}{{(x^{2} + x + 1)}^{2}} d x

Let f (x) = \frac{e^{3 i x}}{{(x^{2} + x + 1)}^{2}} . Poles of f (x) : e^{\frac{2}{3} i π} and e^{- \frac{2}{3} i π}, order - 2

Υ = R e \int_{- \infty}^{+ \infty} f (x) d x = R e (2 i π R e s (f, e^{\frac{2}{3} i π}))

R e s (f, e^{\frac{2}{3} i π}) = \lim_{x \to e^{\frac{2}{3} i π}} {{(x - e^{\frac{2}{3} i π})}^{2} f (x)}^{(1)} = \lim_{x \to e^{\frac{2}{3} i π}} {\frac{e^{3 i x}}{{(x - e^{- \frac{2}{3} i π})}^{2}}}^{(1)}

= \lim_{x \to e^{\frac{2}{3} i π}} {\frac{3 {i e}^{3 i x} {(x - e^{- \frac{2}{3} i π})}^{2} - 2 e^{3 i x} (x - e^{- \frac{2}{3} i π})}{{(x - e^{- \frac{2}{3} i π})}^{4}}}

= - \frac{1}{9} (9 {i e}^{3 i (- \frac{1}{2} + i \frac{\sqrt{3}}{2})} + 2 \sqrt{3} {i e}^{3 i (- \frac{1}{2} + i \frac{\sqrt{3}}{2})}) = - \frac{1}{9} (9 + 2 \sqrt{3}) e^{(- \frac{3 \sqrt{3}}{2} - i \frac{3 - π}{2})}

Υ = - \frac{2 π}{9} (9 + 2 \sqrt{3}) e^{- \frac{3 \sqrt{3}}{2}} \sin (\frac{3 - π}{2}) = \frac{2 π}{9} (9 + 2 \sqrt{3}) e^{- \frac{3 \sqrt{3}}{2}} \cos (\frac{3}{2})

Commented by Mathspace last updated on 26/Dec/21

e r r o r o f c a l c u l u s . .

Commented by Ar Brandon last updated on 26/Dec/21

Thank you for checking, Sir . Rectified!

Answered by Mathspace last updated on 26/Dec/21

Ψ = \int_{- \infty}^{+ \infty} \frac{c o s (3 x)}{{(x^{2} + x + 1)}^{2}} d x \Rightarrow

Ψ = R e (\int_{- \infty}^{+ \infty} \frac{e^{3 i x}}{{(x^{2} + x + 1)}^{2}} d x)

l e t φ (z) = \frac{e^{3 i z}}{{(z^{2} + z + 1)}^{2}} p o l e s o f φ ?

z^{2} + z + 1 = 0 \to Δ = 1 - 4 = - 3 \Rightarrow

z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}}

z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}

t h e p o l e s a r e z_{i} (w i t h o r d r e = 2)

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

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1})

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

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

= {l i m}_{z \to z 1} \frac{3 {i e}^{3 i z} {(z - z_{2})}^{2} - 2 (z - z_{2}) e^{3 i z}}{{(z - z_{2})}^{4}}

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

= \frac{{3 i (z_{1} - z_{2}) - 2} e^{3 {i z}_{1}}}{{(z_{1} - z_{2})}^{3}}

= \frac{{3 i (i \sqrt{3}) - 2} e^{3 i (\frac{- 1 + i \sqrt{3}}{2})}}{{(i \sqrt{3})}^{3}}

= \frac{{- 3 \sqrt{3} - 2} e^{- \frac{3 \sqrt{3}}{2}} {c o s (\frac{3}{2}) - i s i n (\frac{3}{2})}}{- 3 i \sqrt{3}}

= \frac{{3 \sqrt{3} + 2} e^{- \frac{3 \sqrt{3}}{2}} {c o s (\frac{3}{2}) - i s i n (\frac{3}{2})}}{3 i \sqrt{3}}

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

\Rightarrow Ψ = \frac{2 π}{3 \sqrt{3}} (3 \sqrt{3} + 2) e^{- \frac{3 \sqrt{3}}{2}} c o s (\frac{3}{2})