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Question Number 111104 by Lordose last updated on 02/Sep/20

Commented by mathdave last updated on 02/Sep/20

$t h e q u e s t i o n i s n o t i n t e g r a d a b l e$
Commented by Dwaipayan Shikari last updated on 02/Sep/20

$\int {(- 1)}^{x} d x = \int e^{x l o g (- 1)} d x = \int e^{i π x} d x = \frac{1}{i π} e^{i π x} + C$
	Answered by abdomsup last updated on 02/Sep/20

	$t h i s i n t e g r a l c a n b e s o l v e d . . .$
	Commented by mathdave last updated on 02/Sep/20

	$t h e n s o l v e i t s i n c e i t i s i n t e g r a d a b l e$
	Answered by mathmax by abdo last updated on 03/Sep/20
	$I =∫_(−∞) ^(+∞) (((−1)^x^2 )/((x^2 +x+1)^2 )) dx ⇒ I =∫_(−∞) ^(+∞) (e^(iπx^2 ) /((x^2 +x+1)^2 ))dx let ϕ(z) =(e^(iπz^2 ) /((z^2 +z+1)^2 )) poles of ϕ? z^2 +z +1 =0→Δ =1−4 =−3 ⇒z_1 =((−1+i(√3))/2) =e^((i2π)/3) z_2 =((−1−i(√3))/2) =e^(−((i2π)/3)) ⇒ϕ(z) =(e^(iπz^2 ) /((z−e^((i2π)/3) )^2 (z−e^(−((i2π)/3)) )^2 )) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((i2π)/3) ) Res(ϕ ,e^((i2π)/3) ) =lim_(z→e^((i2π)/3) ) (1/((2−1)!)){ (z−e^((i2π)/3) )^2 ϕ(z)}^((1)) =lim_(z→e^((i2π)/3) ) {(e^(iπz^2 ) /((z−e^(−((i2π)/3)) )^2 ))}^((1)) =lim_(z→e^((i2π)/3) ) ((2iπze^(iπz^2 ) (z−e^(−((i2π)/3)) )^2 −2(z−e^(−((i2π)/3)) )e^(iπz^2 ) )/((z−e^(−((i2π)/3)) )^4 )) =lim_(z→e^((i2π)/3) ) ((2iπz e^(iπz^2 ) (z−e^(−((i2π)/3)) )−2e^(iπz^2 ) )/((z−e^(−((i2π)/3)) )^3 )) =lim_(z→e^((i2π)/3) ) ((2e^(iπz^2 ) {iπz (z−e^(−((i2π)/3)) )−1})/((z−e^(−((i2π)/3)) )^3 )) =((2e^(iπe^(i((4π)/3)) ) {iπ e^((i2π)/3) (2isin(((2π)/3))−1})/((2isin(((2π)/3)))^3 )) =((2e^(iπ{−e^((iπ)/3) }) {−π(√3)e^((i2π)/3) −1})/(−8i(((√3)/2))^3 )) =2×((e^(−iπ{(1/2)+((i(√3))/2)}) {π(√3)(−(1/2)+((i(√3))/2)+1})/(3i(√3))) =(2/(3i(√3))) × (−i)e^((π(√3))/2) {((π(√3))/2) +((3π)/2)i} =−(1/(3(√3))) e^((π(√3))/2) {π(√3)+3iπ} ⇒ ∫_(−∞) ^(+∞ ) ϕ(z)dz =2iπ(−(1/(3(√3))))e^((π(√3))/2) {π(√3)+3iπ} =I$
	$I = \int_{- \infty}^{+ \infty} \frac{{(- 1)}^{x^{2}}}{{(x^{2} + x + 1)}^{2}} dx \Rightarrow I = \int_{- \infty}^{+ \infty} \frac{e^{i π x^{2}}}{{(x^{2} + x + 1)}^{2}} dx let$ $φ (z) = \frac{e^{i π z^{2}}}{{(z^{2} + z + 1)}^{2}} poles of φ ?$ $z^{2} + z + 1 = 0 \to Δ = 1 - 4 = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}}$ $z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}} \Rightarrow φ (z) = \frac{e^{i π z^{2}}}{{(z - e^{\frac{i 2 π}{3}})}^{2} {(z - e^{- \frac{i 2 π}{3}})}^{2}}$ $residus theorem give \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, e^{\frac{i 2 π}{3}})$ $Res (φ, e^{\frac{i 2 π}{3}}) = \lim_{z \to e^{\frac{i 2 π}{3}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i 2 π}{3}})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to e^{\frac{i 2 π}{3}}} {\frac{e^{i π z^{2}}}{{(z - e^{- \frac{i 2 π}{3}})}^{2}}}^{(1)}$ $= \lim_{z \to e^{\frac{i 2 π}{3}}} \frac{2 i π {ze}^{i π z^{2}} {(z - e^{- \frac{i 2 π}{3}})}^{2} - 2 (z - e^{- \frac{i 2 π}{3}}) e^{i π z^{2}}}{{(z - e^{- \frac{i 2 π}{3}})}^{4}}$ $= \lim_{z \to e^{\frac{i 2 π}{3}}} \frac{2 i π z e^{i π z^{2}} (z - e^{- \frac{i 2 π}{3}}) - {2 e}^{i π z^{2}}}{{(z - e^{- \frac{i 2 π}{3}})}^{3}}$ $= \lim_{z \to e^{\frac{i 2 π}{3}}} \frac{{2 e}^{i π z^{2}} {i π z (z - e^{- \frac{i 2 π}{3}}) - 1}}{{(z - e^{- \frac{i 2 π}{3}})}^{3}}$ $= \frac{{2 e}^{i π e^{i \frac{4 π}{3}}} {i π e^{\frac{i 2 π}{3}} (2 isin (\frac{2 π}{3}) - 1}}{{(2 isin (\frac{2 π}{3}))}^{3}} = \frac{{2 e}^{i π {- e^{\frac{i π}{3}}}} {- π \sqrt{3} e^{\frac{i 2 π}{3}} - 1}}{- 8 i {(\frac{\sqrt{3}}{2})}^{3}}$ $= 2 \times \frac{e^{- i π {\frac{1}{2} + \frac{i \sqrt{3}}{2}}} {π \sqrt{3} (- \frac{1}{2} + \frac{i \sqrt{3}}{2} + 1}}{3 i \sqrt{3}}$ $= \frac{2}{3 i \sqrt{3}} \times (- i) e^{\frac{π \sqrt{3}}{2}} {\frac{π \sqrt{3}}{2} + \frac{3 π}{2} i} = - \frac{1}{3 \sqrt{3}} e^{\frac{π \sqrt{3}}{2}} {π \sqrt{3} + 3 i π} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π (- \frac{1}{3 \sqrt{3}}) e^{\frac{π \sqrt{3}}{2}} {π \sqrt{3} + 3 i π} = I$
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