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Question Number 79106 by mathmax by abdo last updated on 22/Jan/20
calculate ∫_0 ^∞ e^(−(x^2 +(1/x^2 ))) dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} {dx} \\ $$
Commented by mathmax by abdo last updated on 24/Jan/20
we have proved that ∫_0 ^∞ e^(−(x^2 +(a/x^2 ))) dx =((√π)/2) e^(−2(√a)) for a>0 a=1 ⇒∫_0 ^∞ e^(−(x^2 +(1/x^2 ))) dx =((√π)/2) e^(−2) =((√π)/(2e^2 ))
$${we}\:{have}\:{proved}\:{that}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({x}^{\mathrm{2}} \:+\frac{{a}}{{x}^{\mathrm{2}} }\right)} {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{−\mathrm{2}\sqrt{{a}}} \:\:\:\:{for}\:{a}>\mathrm{0}\: \\ $$$${a}=\mathrm{1}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{−\mathrm{2}} \:=\frac{\sqrt{\pi}}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 24/Jan/20
let I =∫_0 ^∞ e^(−(x^2 +(1/x^2 ))) dx changement x=(1/t) give I =∫_0 ^∞ e^(−(t^2 +(1/t^2 ))) (dt/t^2 ) ⇒2I =∫_0 ^∞ (1+(1/t^2 ))e^(−(t^2 +(1/t^2 ))) dt =∫_0 ^∞ (1+(1/t^2 ))e^(−{(t−(1/t))^2 +2}) dt =e^(−2) ∫_0 ^∞ (1+(1/t^2 ))e^(−(t−(1/t))^2 ) dt =_(t−(1/t)=u) e^(−2) ∫_(−∞) ^(+∞) e^(−u^2 ) du =(√π)e^(−2) =2I ⇒I =((√π)/(2e^2 ))
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} {dx}\:\:{changement}\:{x}=\frac{\mathrm{1}}{{t}}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left({t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} \:\frac{{dt}}{{t}^{\mathrm{2}} }\:\Rightarrow\mathrm{2}{I}\:=\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){e}^{−\left({t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)} \:{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){e}^{−\left\{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}\right\}} {dt}\:={e}^{−\mathrm{2}} \int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){e}^{−\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} } {dt} \\ $$$$=_{{t}−\frac{\mathrm{1}}{{t}}={u}} \:\:\:\:{e}^{−\mathrm{2}} \:\int_{−\infty} ^{+\infty} {e}^{−{u}^{\mathrm{2}} } {du}\:=\sqrt{\pi}{e}^{−\mathrm{2}} \:=\mathrm{2}{I}\:\Rightarrow{I}\:=\frac{\sqrt{\pi}}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$
Answered by mind is power last updated on 22/Jan/20
=∫_0 ^(+∞) e^(−(x^2 +(1/x^2 ))) dx,x=(1/y)⇒∫_0 ^∞ e^(−((1/y^2 )+y^2 )) .(dy/y^2 ) ⇒2∫_0 ^(+∞) e^(−(x^2 +(1/x^2 ))) dx=∫_0 ^(+∞) (1+(1/y^2 ))e^(−(y^2 +(1/y^2 ))) dy ∫_0 ^(+∞) (1+(1/y^2 ))e^(−(y^2 +(1/y^2 ))) dy =∫_0 ^(+∞) (1+(1/y^2 ))e^(−(y−(1/y))^2 −2) =∫_0 ^1 (1+(1/y^2 ))e^(−(y−(1/y))^2 −2) +∫_1 ^(+∞) (1+(1/y^2 ))e^(−(y−(1/y))^2 −2) ∫_0 ^1 (1+(1/y^2 ))e^(−(y−(1/y))^2 −2) dy=,x=y−(1/y)⇒ =∫_(−∞) ^0 e^(−x^2 ) dx=∫_0 ^(+∞) e^(−x^2 −2) dx ∫_1 ^(+∞) (1+(1/y^2 ))e^(−(y−(1/y))^2 −2) =u=(1/y)⇒∫_1 ^0 (1+u^2 )e^(−((1/u)−u)^2 −2) −(du/u^2 ) =∫_0 ^1 (1+(1/u^2 ))e^(−(u−(1/u))^2 −2) du=∫_0 ^(+∞) e^(−x^2 −2) dx ⇒2∫_0 ^(+∞) e^(−(x^2 +(1/x^2 ))) dx=2e^(−2) ∫_0 ^(+∞) e^(−x^2 ) dx=e^(−2) (√π) ⇒∫_0 ^(+∞) e^(−(x^2 +(1/x^2 ))) dx=((√π)/(2e^2 ))
$$=\int_{\mathrm{0}} ^{+\infty} {e}^{−\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} {dx},{x}=\frac{\mathrm{1}}{{y}}\Rightarrow\int_{\mathrm{0}} ^{\infty} {e}^{−\left(\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+{y}^{\mathrm{2}} \right)} .\frac{{dy}}{{y}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {e}^{−\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} {dx}=\int_{\mathrm{0}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\left({y}^{\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)} {dy} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\left({y}^{\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right)} {dy} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}} +\int_{\mathrm{1}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}} {dy}=,{x}={y}−\frac{\mathrm{1}}{{y}}\Rightarrow \\ $$$$=\int_{−\infty} ^{\mathrm{0}} {e}^{−{x}^{\mathrm{2}} } {dx}=\int_{\mathrm{0}} ^{+\infty} {e}^{−{x}^{\mathrm{2}} −\mathrm{2}} {dx} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\right){e}^{−\left({y}−\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} −\mathrm{2}} ={u}=\frac{\mathrm{1}}{{y}}\Rightarrow\int_{\mathrm{1}} ^{\mathrm{0}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right){e}^{−\left(\frac{\mathrm{1}}{{u}}−{u}\right)^{\mathrm{2}} −\mathrm{2}} −\frac{{du}}{{u}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right){e}^{−\left({u}−\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} −\mathrm{2}} {du}=\int_{\mathrm{0}} ^{+\infty} {e}^{−{x}^{\mathrm{2}} −\mathrm{2}} {dx} \\ $$$$\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{+\infty} {e}^{−\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} {dx}=\mathrm{2}{e}^{−\mathrm{2}} \int_{\mathrm{0}} ^{+\infty} {e}^{−{x}^{\mathrm{2}} } {dx}={e}^{−\mathrm{2}} \sqrt{\pi} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} {e}^{−\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} {dx}=\frac{\sqrt{\pi}}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$

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