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Question Number 145724 by Mrsof last updated on 07/Jul/21

Commented by Mrsof last updated on 07/Jul/21

$h o w c a n i g e t t h e r e s u l t o f t h i s e q u a t i o n$
	Answered by phanphuoc last updated on 07/Jul/21
	$1)f(z)=(e^(iz) /((z^2 +a^2 )(z^2 +b^2 ))),∫f(z)=2πi{Res(f(z),ai)+Res(f(z),bi)}=answar....$
	$1) f (z) = \frac{e^{i z}}{(z^{2} + a^{2}) (z^{2} + b^{2})}, \int f (z) = 2 π i {R e s (f (z), a i) + R e s (f (z), b i)} = a n s w a r . . . .$
	Answered by mathmax by abdo last updated on 07/Jul/21
	$1)I=∫_(−∞) ^(+∞) ((cosx)/((x^2 +a^2 )(x^2 +b^2 )))dx ⇒I=Re(∫_(−∞) ^(+∞) (e^(ix) /((x^2 +a^2 )(x^2 +b^2 )))dx) consider ϕ(z)=(e^(iz) /((z^2 +a^2 )(z^2 +b^2 )))=(e^(iz) /((z−ia)(z+ia)(z−ib)(z+ib))) ∫_R ϕ(z)dz =2iπ{Res(ϕ,ia)+Res(ϕ,ib)) Res(ϕ,ia)=(e^(i(ia)) /(2ia(b^2 −a^2 )))=(e^(−a) /(2ia(b^2 −a^2 ))) Res(ϕ,ib)=(e^(i(ib)) /(2ib(a^2 −b^2 ))) =(e^(−b) /(2ib(a^2 −b^2 ))) ∫_R ϕ(z)dz=2iπ{(e^(−a) /(2ia(b^2 −a^2 )))+(e^(−b) /(2ib(a^2 −b^2 )))} =(π/(a^2 −b^2 )){(e^(−b) /b)−(e^(−a) /a)} ⇒I =(π/(a^2 −b^2 )){(e^(−b) /b)−(e^(−a) /a)}$
	$1) I = \int_{- \infty}^{+ \infty} \frac{cosx}{(x^{2} + a^{2}) (x^{2} + b^{2})} dx \Rightarrow I = Re (\int_{- \infty}^{+ \infty} \frac{e^{ix}}{(x^{2} + a^{2}) (x^{2} + b^{2})} dx)$ $consider φ (z) = \frac{e^{iz}}{(z^{2} + a^{2}) (z^{2} + b^{2})} = \frac{e^{iz}}{(z - ia) (z + ia) (z - ib) (z + ib)}$ $\int_{R} φ (z) dz = 2 i π {Res (φ, ia) + Res (φ, ib))$ $Res (φ, ia) = \frac{e^{i (ia)}}{2 ia (b^{2} - a^{2})} = \frac{e^{- a}}{2 ia (b^{2} - a^{2})}$ $Res (φ, ib) = \frac{e^{i (ib)}}{2 ib (a^{2} - b^{2})} = \frac{e^{- b}}{2 ib (a^{2} - b^{2})}$ $\int_{R} φ (z) dz = 2 i π {\frac{e^{- a}}{2 ia (b^{2} - a^{2})} + \frac{e^{- b}}{2 ib (a^{2} - b^{2})}}$ $= \frac{π}{a^{2} - b^{2}} {\frac{e^{- b}}{b} - \frac{e^{- a}}{a}} \Rightarrow I = \frac{π}{a^{2} - b^{2}} {\frac{e^{- b}}{b} - \frac{e^{- a}}{a}}$
	Answered by mathmax by abdo last updated on 07/Jul/21

	$2) J = \int_{0}^{\infty} \frac{\cos (ax)}{x^{2} + 1} dx \Rightarrow 2 J = \int_{- \infty}^{+ \infty} \frac{\cos (ax)}{x^{2} + 1} dx = Re (\int_{- \infty}^{+ \infty} \frac{e^{iax}}{x^{2} + 1} dx)$ $and \int_{- \infty}^{+ \infty} \frac{e^{iax}}{x^{2} + 1} dx = 2 i π Res (φ, i) with φ (z) = \frac{e^{iaz}}{z^{2} + 1} = \frac{e^{iaz}}{(z - i) (z + i)}$ $Res (φ, i) = \frac{e^{iai}}{2 i} = \frac{e^{- a}}{2 i} \Rightarrow \int_{R} φ (z) dz = 2 i π \frac{e^{- a}}{2 i} = π e^{- a} \Rightarrow$ $J = \frac{π}{2} e^{- a}$
	Answered by mathmax by abdo last updated on 07/Jul/21
	$3)Υ(b)=∫_0 ^∞ ((cos(ax))/(x^2 +b^2 ))dx ⇒Υ^′ (b)=−2b∫_0 ^∞ ((cos(ax))/((x^2 +b^2 )^2 ))dx ⇒ ∫_0 ^∞ ((cos(ax))/((x^2 +b^2 )^2 ))=−(1/(2b))Υ^′ (b) we have 2Υ(b)=Re(∫_(−∞) ^(+∞) (e^(iax) /(x^2 +b^2 ))dx) ∫_(−∞) ^(+∞) (e^(iax) /(x^2 +b^2 ))dx=2iπRes(ϕ,ib)=2iπ×(e^(ia(ib)) /(2ib))=(π/b)e^(−ab) ⇒ Υ(b)=(π/(2b))e^(−ab) ⇒Υ^′ (b)=(π/2){−(1/b^2 )e^(−ab) −(a/b)e^(−ab) } =−(π/(2b^2 ))(1+ab)e^(−ab) ⇒ ∫_0 ^∞ ((cos(ax))/((x^2 +b^2 )^2 ))=(−(1/(2b)))(−(π/(2b^2 )))(1+ab)e^(−ab) =(π/(4b^3 ))(1+ab)e^(−ab)$
	$3) Υ (b) = \int_{0}^{\infty} \frac{\cos (ax)}{x^{2} + b^{2}} dx \Rightarrow Υ^{^{'}} (b) = - 2 b \int_{0}^{\infty} \frac{\cos (ax)}{{(x^{2} + b^{2})}^{2}} dx \Rightarrow$ $\int_{0}^{\infty} \frac{\cos (ax)}{{(x^{2} + b^{2})}^{2}} = - \frac{1}{2 b} Υ^{^{'}} (b) we have 2 Υ (b) = Re (\int_{- \infty}^{+ \infty} \frac{e^{iax}}{x^{2} + b^{2}} dx)$ $\int_{- \infty}^{+ \infty} \frac{e^{iax}}{x^{2} + b^{2}} dx = 2 i π Res (φ, ib) = 2 i π \times \frac{e^{ia (ib)}}{2 ib} = \frac{π}{b} e^{- ab} \Rightarrow$ $Υ (b) = \frac{π}{2 b} e^{- ab} \Rightarrow Υ^{^{'}} (b) = \frac{π}{2} {- \frac{1}{b^{2}} e^{- ab} - \frac{a}{b} e^{- ab}}$ $= - \frac{π}{{2 b}^{2}} (1 + ab) e^{- ab} \Rightarrow$ $\int_{0}^{\infty} \frac{\cos (ax)}{{(x^{2} + b^{2})}^{2}} = (- \frac{1}{2 b}) (- \frac{π}{{2 b}^{2}}) (1 + ab) e^{- ab}$ $= \frac{π}{{4 b}^{3}} (1 + ab) e^{- ab}$
	Answered by mathmax by abdo last updated on 07/Jul/21

	$4) Ψ = \int_{0}^{\infty} \frac{xsin (2 x)}{x^{2} + 3} dx \Rightarrow 2 Ψ = \int_{- \infty}^{+ \infty} \frac{xsin (2 x)}{x^{2} + 3} dx$ $= Im (\int_{- \infty}^{+ \infty} \frac{{xe}^{2 ix}}{x^{2} + 3} dx) let Λ (z) = \frac{{ze}^{2 iz}}{z^{2} + 3} \Rightarrow Λ (z) = \frac{{ze}^{2 iz}}{(z - i \sqrt{3}) (z + i \sqrt{3})}$ $\int_{R} Λ (z) dz = 2 i π Res (Λ, i \sqrt{3}) = 2 i π \times \frac{i \sqrt{3} e^{2 i (i \sqrt{3})}}{2 i \sqrt{3}} = i π e^{- 2 \sqrt{3}}$ $Ψ = \frac{π}{2} e^{- 2 \sqrt{3}}$
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