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Question Number 84693 by Power last updated on 15/Mar/20
Commented by abdomathmax last updated on 15/Mar/20
let ϕ(t)=∫_0 ^∞ ((sin(tx))/(x(x^2 +1)))dx we have ϕ^′ (t)=∫_0 ^∞ ((cos(tx))/(x^2 +1))dx ⇒2ϕ^′ (t)=∫_(−∞) ^(+∞) ((cos(tx))/(x^2 +1))dx =Re(∫_(−∞) ^(+∞) (e^(itx) /(x^2 +1))dx) residus theorem give ∫_(−∞) ^(+∞) (e^(itx) /(x^2 +1))dx =2iπ Res(f,i) =2iπ×(e^(−t) /(2i)) =πe^(−t) ⇒2ϕ^′ (t)=πe^(−t) ⇒ϕ^′ (t)=(π/2)e^(−t) ⇒ ϕ(t)=k−(π/2)e^(−t) ϕ(0)=0=k−(π/2) ⇒k=(π/2) ⇒ϕ(t)=(π/2)−(π/2)e^(−t) I =∫_0 ^∞ ((sinx)/(x(1+x^2 )))dx =ϕ(1)=(π/2)−(π/(2e))
$${let}\:\varphi\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({tx}\right)}{{x}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}{dx}\:\:{we}\:{have} \\ $$$$\varphi^{'} \left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({tx}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:\Rightarrow\mathrm{2}\varphi^{'} \left({t}\right)=\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({tx}\right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$={Re}\left(\int_{−\infty} ^{+\infty} \:\frac{{e}^{{itx}} }{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\right)\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\frac{{e}^{{itx}} }{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:=\mathrm{2}{i}\pi\:{Res}\left({f},{i}\right)\:=\mathrm{2}{i}\pi×\frac{{e}^{−{t}} }{\mathrm{2}{i}} \\ $$$$=\pi{e}^{−{t}} \:\Rightarrow\mathrm{2}\varphi^{'} \left({t}\right)=\pi{e}^{−{t}} \:\Rightarrow\varphi^{'} \left({t}\right)=\frac{\pi}{\mathrm{2}}{e}^{−{t}} \:\Rightarrow \\ $$$$\varphi\left({t}\right)={k}−\frac{\pi}{\mathrm{2}}{e}^{−{t}} \\ $$$$\varphi\left(\mathrm{0}\right)=\mathrm{0}={k}−\frac{\pi}{\mathrm{2}}\:\Rightarrow{k}=\frac{\pi}{\mathrm{2}}\:\Rightarrow\varphi\left({t}\right)=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}{e}^{−{t}} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\:=\varphi\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}{e}} \\ $$
Answered by Joel578 last updated on 15/Mar/20
I(t) = ∫_0 ^∞ ((sin (tx))/(x(1 + x^2 ))) dx ⇒ I ′(t) = ∫_0 ^∞ ((cos (tx))/(1 + x^2 )) dx ⇒ I ′′(t) = −∫_0 ^∞ ((x sin (tx))/(1 + x^2 )) dx = −∫_0 ^∞ (((x^2 + 1 − 1)sin (tx))/(x(1 + x^2 ))) dx = −∫_0 ^∞ ((sin (tx))/x) dx + ∫_0 ^∞ ((sin (tx))/(x(1 + x^2 ))) dx ⇒ I ′′(t) = −(π/2) + I(t) ⇒ I ′′ − I = −(π/2) Solving non−homogeneous ODE gives ⇒ I(t) = C_1 e^t + C_2 e^(−t) + (π/2) ⇒ I ′(t) = C_1 e^t − C_2 e^(−t) Note that I(0) = ∫_0 ^∞ 0 dx = C_1 + C_2 + (π/2) = 0 and I ′(0) = ∫_0 ^∞ (1/(1 + x^2 )) dx = C_1 − C_2 = (π/2) ⇒ C_1 = 0, C_2 = −(π/2) ⇒ I(t) = ∫_0 ^∞ ((sin (tx))/(x(1 + x^2 ))) dx = −(π/2)e^(−t) + (π/2) ⇒ I(1) = ∫_0 ^∞ ((sin (x))/(x(1 + x^2 ))) dx = −(π/(2e)) + (π/2)
$${I}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\:\left({tx}\right)}{{x}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}\:{dx} \\ $$$$\Rightarrow\:{I}\:'\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cos}\:\left({tx}\right)}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx} \\ $$$$\Rightarrow\:{I}\:''\left({t}\right)\:=\:−\int_{\mathrm{0}} ^{\infty} \:\frac{{x}\:\mathrm{sin}\:\left({tx}\right)}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx}\:=\:\:−\int_{\mathrm{0}} ^{\infty} \:\frac{\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\:−\:\mathrm{1}\right)\mathrm{sin}\:\left({tx}\right)}{{x}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\:\left({tx}\right)}{{x}}\:{dx}\:+\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\:\left({tx}\right)}{{x}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}\:{dx} \\ $$$$\Rightarrow\:{I}\:''\left({t}\right)\:=\:−\frac{\pi}{\mathrm{2}}\:+\:{I}\left({t}\right) \\ $$$$\Rightarrow\:{I}\:''\:−\:{I}\:=\:−\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{Solving}\:\mathrm{non}−\mathrm{homogeneous}\:\mathrm{ODE}\:\mathrm{gives} \\ $$$$\Rightarrow\:{I}\left({t}\right)\:=\:{C}_{\mathrm{1}} {e}^{{t}} \:+\:{C}_{\mathrm{2}} {e}^{−{t}} \:+\:\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:{I}\:'\left({t}\right)\:=\:{C}_{\mathrm{1}} {e}^{{t}} \:−\:{C}_{\mathrm{2}} {e}^{−{t}} \\ $$$$\mathrm{Note}\:\mathrm{that}\:{I}\left(\mathrm{0}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \mathrm{0}\:{dx}\:=\:{C}_{\mathrm{1}} \:+\:{C}_{\mathrm{2}} \:+\:\frac{\pi}{\mathrm{2}}\:=\:\mathrm{0}\: \\ $$$$\mathrm{and}\:{I}\:'\left(\mathrm{0}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}\:+\:{x}^{\mathrm{2}} }\:{dx}\:=\:{C}_{\mathrm{1}} \:−\:{C}_{\mathrm{2}} \:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:{C}_{\mathrm{1}} \:=\:\mathrm{0},\:{C}_{\mathrm{2}} \:=\:−\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:{I}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:\left({tx}\right)}{{x}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}\:{dx}\:=\:−\frac{\pi}{\mathrm{2}}{e}^{−{t}} \:+\:\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\:{I}\left(\mathrm{1}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\:\left({x}\right)}{{x}\left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)}\:{dx}\:=\:−\frac{\pi}{\mathrm{2}{e}}\:+\:\frac{\pi}{\mathrm{2}} \\ $$
Commented by Power last updated on 15/Mar/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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