find ∫_(−∞) ^(+∞) ((cos(2x^2 +1))/(x^4 −x^2 +3))dx


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Question Number 80452 by abdomathmax last updated on 03/Feb/20

$f i n d \int_{- \infty}^{+ \infty} \frac{c o s (2 x^{2} + 1)}{x^{4} - x^{2} + 3} d x$
Commented by abdomathmax last updated on 03/Mar/20
$I= Re(∫_(−∞) ^(+∞) (e^(i(2x^2 +1)) /(x^4 −x^2 +3)) dx)let W(z)=(e^(i(2z^2 +1)) /(z^4 −z^2 +3)) poles of W? z^4 −z^(2 ) +3=0 ⇒t^2 −t +3=0 (t=z^2 ) Δ=1−12=−11 ⇒t_1 =((1+i(√(11)))/2) t_2 =((1−i(√(11)))/2) ⇒W(z) =(e^(i(2z^2 +1)) /((z^2 −t_1 )(z^2 −t_2 ))) =(e^(i(2z^2 +1)) /((z−(√t_1 ))(z+(√t_1 ))(z−(√t_2 ))(z+(√t_2 )))) ∣t_1 ∣=(1/2)(√(12))=(1/2)(2(√3)) =(√3) ⇒t_1 =(√3)e^(iarctan((√(11)))) t_2 =(√3)e^(−isrcran((√(11)))) ∫_(−∞) ^(+∞ ) W(z)dz =2iπ {Res(W,(√t_1 )) +Res(W,−(√t_2 ))} =2iπ {Res(W,^4 (√3)e^((i/2)arcran((√(11)))) )+Res(W,−^4 (√3)e^(−(i/2)arcran((√(11)))) } Res(W,(√t_1 )) =(e^(i(2t_1 +1)) /(2(√t_1 )×(t_1 −t_2 ))) Res(W,−(√t_2 )) =(e^(i(2t_2 +1)) /(−2(√t_2 )×(t_2 −t_1 ))) ⇒ ∫_(−∞) ^(+∞) W(z)dz =((iπ)/(t_1 −t_2 )){ (e^(i(2t_1 +1)) /(√t_1 )) +(e^(i(2t_2 +1)) /(√t_2 ))} ....be continued....$
$I = R e (\int_{- \infty}^{+ \infty} \frac{e^{i (2 x^{2} + 1)}}{x^{4} - x^{2} + 3} d x) l e t W (z) = \frac{e^{i (2 z^{2} + 1)}}{z^{4} - z^{2} + 3}$ $p o l e s o f W ?$ $z^{4} - z^{2} + 3 = 0 \Rightarrow t^{2} - t + 3 = 0 (t = z^{2})$ $Δ = 1 - 12 = - 11 \Rightarrow t_{1} = \frac{1 + i \sqrt{11}}{2}$ $t_{2} = \frac{1 - i \sqrt{11}}{2} \Rightarrow W (z) = \frac{e^{i (2 z^{2} + 1)}}{(z^{2} - t_{1}) (z^{2} - t_{2})}$ $= \frac{e^{i (2 z^{2} + 1)}}{(z - \sqrt{t_{1}}) (z + \sqrt{t_{1}}) (z - \sqrt{t_{2}}) (z + \sqrt{t_{2}})}$ $∣ t_{1} ∣= \frac{1}{2} \sqrt{12} = \frac{1}{2} (2 \sqrt{3}) = \sqrt{3} \Rightarrow t_{1} = \sqrt{3} e^{i a r c t a n (\sqrt{11})}$ $t_{2} = \sqrt{3} e^{- i s r c r a n (\sqrt{11})}$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π {R e s (W, \sqrt{t_{1}}) + R e s (W, - \sqrt{t_{2}})}$ $= 2 i π {R e s (W,^{4} \sqrt{3} e^{\frac{i}{2} a r c r a n (\sqrt{11})}) + R e s (W, -^{4} \sqrt{3} e^{- \frac{i}{2} a r c r a n (\sqrt{11})}}$ $R e s (W, \sqrt{t_{1}}) = \frac{e^{i (2 t_{1} + 1)}}{2 \sqrt{t_{1}} \times (\underset{1}{t} - t_{2})}$ $R e s (W, - \sqrt{t_{2}}) = \frac{e^{i (2 t_{2} + 1)}}{- 2 \sqrt{t_{2}} \times (t_{2} - t_{1})} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = \frac{i π}{t_{1} - t_{2}} {\frac{e^{i (2 t_{1} + 1)}}{\sqrt{t_{1}}} + \frac{e^{i (2 t_{2} + 1)}}{\sqrt{t_{2}}}}$ $. . . . b e c o n t i n u e d . . . .$
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