Question-130326

Question Number 130326 by rs4089 last updated on 24/Jan/21

Answered by Lordose last updated on 24/Jan/21

Ω(p) = ∫_0 ^( ∞) ((sin(px))/(x(x^2 +1)))dx Ω′(p) = ∫_0 ^( ∞) ((cos(px))/(x^2 +1))dx = (π/2)e^(−p) Ω(p) = (π/2)e^(−p) + C Ω(0) = (π/2) = C Ω(p) = (π/2)e^(−p) +(π/2)

$$ \\ $$$$\Omega\left(\mathrm{p}\right)\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sin}\left(\mathrm{px}\right)}{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dx} \\ $$$$\Omega'\left(\mathrm{p}\right)\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{cos}\left(\mathrm{px}\right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mathrm{dx}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{p}} \\ $$$$\Omega\left(\mathrm{p}\right)\:=\:\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{p}} \:+\:\mathrm{C} \\ $$$$\Omega\left(\mathrm{0}\right)\:=\:\frac{\pi}{\mathrm{2}}\:=\:\mathrm{C} \\ $$$$\Omega\left(\mathrm{p}\right)\:=\:\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{p}} +\frac{\pi}{\mathrm{2}} \\ $$

Answered by mathmax by abdo last updated on 24/Jan/21

let f(p)=∫_0 ^∞ ((sin(px))/(x(x^2 +1)))dx ⇒f^′ (p)=∫_0 ^∞ ((cos(px))/(x^2 +1))dx =(1/2)∫_(−∞) ^(+∞) ((cos(px))/(x^2 +1))dx =(1/2)Re(∫_R (e^(ipx) /(x^2 +1))dx) =(1/2)Re(2iπ×(e^(−p) /(2i))) =(1/2)Re(πe^(−p) ) =(π/2)e^(−p) ⇒ f(p)=(π/2)e^(−p) +C f(0)=0 =(π/2) +C ⇒C=−(π/2) ⇒f(p)=(π/2)e^(−p) −(π/2)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{p}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{px}\right)}{\mathrm{x}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\mathrm{dx}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{p}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\left(\mathrm{px}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{px}\right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\int_{\mathrm{R}} \frac{\mathrm{e}^{\mathrm{ipx}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\mathrm{2i}\pi×\frac{\mathrm{e}^{−\mathrm{p}} }{\mathrm{2i}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Re}\left(\pi\mathrm{e}^{−\mathrm{p}} \right)\:=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{p}} \:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{p}\right)=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{p}} \:+\mathrm{C} \\ $$$$\mathrm{f}\left(\mathrm{0}\right)=\mathrm{0}\:=\frac{\pi}{\mathrm{2}}\:+\mathrm{C}\:\Rightarrow\mathrm{C}=−\frac{\pi}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{p}\right)=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mathrm{p}} −\frac{\pi}{\mathrm{2}} \\ $$

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