calculte ∫_(−∞) ^∞ ((sin(πx^2 ))/((x^2 +2x+2)^2 ))dx


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Question Number 137536 by Mathspace last updated on 03/Apr/21

$c a l c u l t e \int_{- \infty}^{\infty} \frac{s i n (π x^{2})}{{(x^{2} + 2 x + 2)}^{2}} d x$
	Answered by mathmax by abdo last updated on 04/Apr/21
	$let f(a)=∫_(−∞) ^(+∞) ((sin(πx^2 ))/(x^2 +2x+a))dx with a>1 we have f^′ (a)=−∫_(−∞) ^(+∞) ((sin(πx^2 ))/((x^2 +2x+a)^2 )) ⇒f^′ (2)=−∫_(−∞) ^(+∞) ((sin(πx^2 ))/((x^2 +2x+2)^2 )) ⇒ ∫_(−∞) ^(+∞) ((sin(πx^2 ))/((x^2 +2x+2)^2 ))=−f^′ (2) we have f(a)=Im(∫_(−∞) ^(+∞) (e^(iπx^2 ) /(x^2 +2x+a))dx) let ϕ(z)=(e^(iπz^2 ) /(z^2 +2z+a)) poles of ϕ? Δ^′ =1−a<0 ⇒z_1 =−1+i(√(a−1)) and z_2 =−1−i(√(a−1)) ϕ(z)=(e^(iπz^2 ) /((z−z_1 )(z−z_2 ))) residus theorem ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) =2iπ×(e^(iπz_1 ^2 ) /(2i(√(a−1)))) =(π/( (√(a−1)))) e^(iπ{1−2i(√(a−1))−a+1)) =(π/( (√(a−1)))) e^(iπ{2−a−2i(√(a−1))}) =(π/( (√(a−1)))) e^(iπ(2−a)) .e^(2π(√(a−1))) =((πe^(2π(√(a−1))) )/( (√(a−1)))){cos(π(2−a))+isin(π(2−a))} ⇒ f(a)=−((π e^(2π(√(a−1))) )/( (√(a−1)))).sin(πa) ⇒ −f^′ (a) =π((e^(2π(√(a−1))) /( (√(a−1)))))^, sin(πa) +π^2 cos(πa).(e^(2π(√(a−1))) /( (√(a−1)))) but ((e^(2π(√(a−1))) /( (√(a−1)))))^′ =((((2π)/(2(√(a−1))))e^(2π(√(a−1))) .(√(a−1))−e^(2π(√(a−1))) .(1/(2(√(a−1)))))/(a−1)) =((πe^(2π(√(a−1))) −(1/(2(√(a−1))))e^(2π(√(a−1))) )/(a−1)) =(((2π(√(a−1))−1)e^(2π(√(a−1))) )/(2(a−1)(√(a−1)))) ⇒ f^′ (a)=−(π/2)(((2π(√(a−1))−1)e^(2π(√(a−1))) )/((a−1)(√(a−1)))) sin(πa) −π^2 cos(πa).(e^(2π(√(a−1))) /( (√(a−1)))) ⇒f^′ (2)=−(π/2)(((2π−1)e^(2π) )/1).0 −π^2 e^(2π) =−π^2 e^(2π) ⇒ ∫_(−∞) ^(+∞) ((sin(πx^2 ))/((x^2 +2x+2)^2 )) =π^2 e^(2π)$
	$let f (a) = \int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{x^{2} + 2 x + a} dx with a > 1$ $we have f^{^{'}} (a) = - \int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{{(x^{2} + 2 x + a)}^{2}} \Rightarrow f^{^{'}} (2) = - \int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{{(x^{2} + 2 x + 2)}^{2}} \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{{(x^{2} + 2 x + 2)}^{2}} = - f^{^{'}} (2)$ $we have f (a) = Im (\int_{- \infty}^{+ \infty} \frac{e^{i π x^{2}}}{x^{2} + 2 x + a} dx) let φ (z) = \frac{e^{i π z^{2}}}{z^{2} + 2 z + a}$ $poles of φ ? Δ^{^{'}} = 1 - a < 0 \Rightarrow z_{1} = - 1 + i \sqrt{a - 1} and z_{2} = - 1 - i \sqrt{a - 1}$ $φ (z) = \frac{e^{i π z^{2}}}{(z - z_{1}) (z - z_{2})} residus theorem \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, z_{1}) = 2 i π \times \frac{e^{i π z_{1}^{2}}}{2 i \sqrt{a - 1}}$ $= \frac{π}{\sqrt{a - 1}} e^{i π {1 - 2 i \sqrt{a - 1} - a + 1)} = \frac{π}{\sqrt{a - 1}} e^{i π {2 - a - 2 i \sqrt{a - 1}}}$ $= \frac{π}{\sqrt{a - 1}} e^{i π (2 - a)} . e^{2 π \sqrt{a - 1}} = \frac{π e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}} {\cos (π (2 - a)) + isin (π (2 - a))} \Rightarrow$ $f (a) = - \frac{π e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}} . \sin (π a) \Rightarrow$ $- f^{^{'}} (a) = π {(\frac{e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}})}^{,} \sin (π a) + π^{2} \cos (π a) . \frac{e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}} but$ ${(\frac{e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}})}^{^{'}} = \frac{\frac{2 π}{2 \sqrt{a - 1}} e^{2 π \sqrt{a - 1}} . \sqrt{a - 1} - e^{2 π \sqrt{a - 1}} . \frac{1}{2 \sqrt{a - 1}}}{a - 1}$ $= \frac{π e^{2 π \sqrt{a - 1}} - \frac{1}{2 \sqrt{a - 1}} e^{2 π \sqrt{a - 1}}}{a - 1} = \frac{(2 π \sqrt{a - 1} - 1) e^{2 π \sqrt{a - 1}}}{2 (a - 1) \sqrt{a - 1}} \Rightarrow$ $f^{^{'}} (a) = - \frac{π}{2} \frac{(2 π \sqrt{a - 1} - 1) e^{2 π \sqrt{a - 1}}}{(a - 1) \sqrt{a - 1}} \sin (π a) - π^{2} \cos (π a) . \frac{e^{2 π \sqrt{a - 1}}}{\sqrt{a - 1}}$ $\Rightarrow f^{^{'}} (2) = - \frac{π}{2} \frac{(2 π - 1) e^{2 π}}{1} . 0 - π^{2} e^{2 π} = - π^{2} e^{2 π} \Rightarrow$ $\int_{- \infty}^{+ \infty} \frac{\sin (π x^{2})}{{(x^{2} + 2 x + 2)}^{2}} = π^{2} e^{2 π}$
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