Menu Close

0-e-x-2-x-2-1-2-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 141719 by qaz last updated on 22/May/21
∫_0 ^∞ (e^(−x^2 ) /((x^2 +(1/2))^2 ))dx=?
$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}=? \\ $$
Answered by MJS_new last updated on 23/May/21
∫(e^(−x^2 ) /((x^2 +(1/2))^2 ))dx=2∫e^(−x^2 ) −2∫(((4x^4 +4x^2 −1)e^(−x^2 ) )/((2x^2 +1)^2 ))dx 2∫e^(−x^2 ) dx=(√π)erf x I tried this with f(x) being a polynome ∫(((4x^4 +4x^2 −1)e^(−x^2 ) )/((2x^2 +1)^2 ))dx=((f(x)e^(−x^2 ) )/(2x^2 +1)) (d/dx)[((f(x)e^(−x^2 ) )/(2x^2 +1))]=(((2x^2 +1)f ′(x)−2x(2x^2 +3)f(x))/((2x^2 +1)^2 ))e^(−x^2 ) ⇒ (2x^2 +1)f ′(x)−2x(2x^2 +3)f(x)=4x^4 +4x^2 −1 good luck! f(x)=−x ⇒ ∫(((4x^4 +4x^2 −1)e^(−x^2 ) )/((2x^2 +1)^2 ))dx=−((xe^(−x^2 ) )/(2x^2 +1)) ⇒ ∫(e^(−x^2 ) /((x^2 +(1/2))^2 ))dx=(√π)erf x +((2xe^(−x^2 ) )/(2x^2 +1))+C ⇒ answer is (√π)
$$\int\frac{\mathrm{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}=\mathrm{2}\int\mathrm{e}^{−{x}^{\mathrm{2}} } −\mathrm{2}\int\frac{\left(\mathrm{4}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{e}^{−{x}^{\mathrm{2}} } }{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{2}\int\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}=\sqrt{\pi}\mathrm{erf}\:{x} \\ $$$$\mathrm{I}\:\mathrm{tried}\:\mathrm{this}\:\mathrm{with}\:{f}\left({x}\right)\:\mathrm{being}\:\mathrm{a}\:\mathrm{polynome} \\ $$$$\int\frac{\left(\mathrm{4}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{e}^{−{x}^{\mathrm{2}} } }{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}=\frac{{f}\left({x}\right)\mathrm{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\frac{{d}}{{dx}}\left[\frac{{f}\left({x}\right)\mathrm{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}\right]=\frac{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right){f}\:'\left({x}\right)−\mathrm{2}{x}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right){f}\left({x}\right)}{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{e}^{−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right){f}\:'\left({x}\right)−\mathrm{2}{x}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right){f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{good}\:\mathrm{luck}! \\ $$$${f}\left({x}\right)=−{x} \\ $$$$\Rightarrow \\ $$$$\int\frac{\left(\mathrm{4}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{e}^{−{x}^{\mathrm{2}} } }{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dx}=−\frac{{x}\mathrm{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow \\ $$$$\int\frac{\mathrm{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx}=\sqrt{\pi}\mathrm{erf}\:{x}\:+\frac{\mathrm{2}{x}\mathrm{e}^{−{x}^{\mathrm{2}} } }{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\sqrt{\pi} \\ $$
Commented by qaz last updated on 23/May/21
Congratulation Sir . You have passed the China Postgraduate Entrance Examination.
$${Congratulation}\:{Sir}\:. \\ $$$${You}\:{have}\:{passed}\:{the}\:{China}\:{Postgraduate}\:{Entrance}\:{Examination}. \\ $$
Commented by greg_ed last updated on 23/May/21
cool !
$$\mathrm{cool}\:! \\ $$
Answered by mathmax by abdo last updated on 23/May/21
Ψ=∫_0 ^∞ (e^(−x^2 ) /((x^2 +(1/2))^2 ))dx =Φ((1/( (√2)))) with Φ(a)=∫_0 ^∞ (e^(−x^2 ) /((x^(2 ) +a^2 )^2 ))dx (a>0) Φ(a)=(1/2)∫_(−∞) ^(+∞) (e^(−x^2 ) /((x^2 +a^2 )^2 ))dx =(1/2)(2iπ)Res(f ,ai) =iπRes(f,ai) f(z)=(e^(−z^2 ) /((z^2 +a^2 )^2 )) ⇒f(z)=(e^(−z^2 ) /((z−ia)^2 (z+ia)^2 )) Res(f,ai) =lim_(z→ai) (1/((2−1)!)){ (e^(−z^2 ) /((z+ia)^2 ))}^((1)) =lim_(z→ai) ((−2ze^(−z^2 ) (z+ia)^2 −2(z+ia)e^(−z^2 ) )/((z+ia)^4 )) =lim_(z→ai) (((−2z (z+ia)−2)e^(−z^2 ) )/((z+ia)^3 )) =−2lim_(z→ia) (((z^2 +iaz +1)e^(−z^2 ) )/((z+ia)^3 )) =−2 (((−a^2 −a^2 +1)e^a^2 )/((2ia)^3 )) =−2 (((1−2a^2 )e^a^2 )/(−8i a^3 )) =(((1−2a^2 )e^a^2 )/(4ia^3 )) ⇒Φ(a)=iπ.(((1−2a^2 )e^a^2 )/(4ia^3 ))=(π/(4a^3 ))(1−2a^2 )e^a^2 ⇒Φ((1/( (√2))))=0 =Ψ...!!
$$\Psi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } }{\left(\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\mathrm{dx}\:=\Phi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:\mathrm{with}\:\Phi\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } }{\left(\mathrm{x}^{\mathrm{2}\:} +\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:\:\left(\mathrm{a}>\mathrm{0}\right) \\ $$$$\Phi\left(\mathrm{a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } }{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2i}\pi\right)\mathrm{Res}\left(\mathrm{f}\:\:,\mathrm{ai}\right)\:=\mathrm{i}\pi\mathrm{Res}\left(\mathrm{f},\mathrm{ai}\right) \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}−\mathrm{ia}\right)^{\mathrm{2}} \left(\mathrm{z}+\mathrm{ia}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{ai}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{ai}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\frac{\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{ia}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{ai}} \:\frac{−\mathrm{2ze}^{−\mathrm{z}^{\mathrm{2}} } \left(\mathrm{z}+\mathrm{ia}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{z}+\mathrm{ia}\right)\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{ia}\right)^{\mathrm{4}} } \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{ai}} \:\:\:\:\:\frac{\left(−\mathrm{2z}\:\left(\mathrm{z}+\mathrm{ia}\right)−\mathrm{2}\right)\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{ia}\right)^{\mathrm{3}} } \\ $$$$=−\mathrm{2lim}_{\mathrm{z}\rightarrow\mathrm{ia}} \:\:\:\frac{\left(\mathrm{z}^{\mathrm{2}} +\mathrm{iaz}\:+\mathrm{1}\right)\mathrm{e}^{−\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}+\mathrm{ia}\right)^{\mathrm{3}} } \\ $$$$=−\mathrm{2}\:\frac{\left(−\mathrm{a}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{e}^{\mathrm{a}^{\mathrm{2}} } }{\left(\mathrm{2ia}\right)^{\mathrm{3}} }\:=−\mathrm{2}\:\frac{\left(\mathrm{1}−\mathrm{2a}^{\mathrm{2}} \right)\mathrm{e}^{\mathrm{a}^{\mathrm{2}} } }{−\mathrm{8i}\:\mathrm{a}^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{1}−\mathrm{2a}^{\mathrm{2}} \right)\mathrm{e}^{\mathrm{a}^{\mathrm{2}} } }{\mathrm{4ia}^{\mathrm{3}} }\:\Rightarrow\Phi\left(\mathrm{a}\right)=\mathrm{i}\pi.\frac{\left(\mathrm{1}−\mathrm{2a}^{\mathrm{2}} \right)\mathrm{e}^{\mathrm{a}^{\mathrm{2}} } }{\mathrm{4ia}^{\mathrm{3}} }=\frac{\pi}{\mathrm{4a}^{\mathrm{3}} }\left(\mathrm{1}−\mathrm{2a}^{\mathrm{2}} \right)\mathrm{e}^{\mathrm{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\Phi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\mathrm{0}\:=\Psi…!! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *