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Question Number 65676 by mathmax by abdo last updated on 01/Aug/19
calculate ∫_0 ^∞ e^(−x^2 −(1/x^2 )) dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}^{\mathrm{2}} −\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} {dx} \\ $$
Commented by ~ À ® @ 237 ~ last updated on 04/Aug/19
let named that integral I . changes x=(1/u) dx=((−du)/u^2 ) I=∫_0 ^∞ e^(−(1/u^2 )−u^2 ) (du/u^2 ) So 2I=∫_0 ^∞ (1+(1/u^2 ))e^(−u^2 −(1/u^2 )) du u^2 +(1/u^2 )=(u−(1/u))^2 +2 then 2I=e^(−2) ∫_0 ^∞ e^(−(u−(1/u))^2 ) (1+(1/u^2 ))du now let change v=(u−(1/u)) dv=(1+(1/u^2 ))du so 2I= e^(−2) ∫_(−∞) ^∞ e^(−v^2 ) dv=e^(−2) (√π) finally I= ((√π)/(2e^2 ))
$$ \\ $$$$ \\ $$$${let}\:{named}\:{that}\:{integral}\:{I}\:.\:{changes}\:\:{x}=\frac{\mathrm{1}}{{u}}\:\:\:{dx}=\frac{−{du}}{{u}^{\mathrm{2}} }\: \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }−{u}^{\mathrm{2}} } \frac{{du}}{{u}^{\mathrm{2}} }\: \\ $$$${So}\:\mathrm{2}{I}=\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right){e}^{−{u}^{\mathrm{2}} −\frac{\mathrm{1}}{{u}^{\mathrm{2}} }} {du} \\ $$$${u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }=\left({u}−\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} +\mathrm{2} \\ $$$${then}\:\:\mathrm{2}{I}={e}^{−\mathrm{2}} \int_{\mathrm{0}} ^{\infty} {e}^{−\left({u}−\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} } \left(\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right){du} \\ $$$${now}\:{let}\:{change}\:{v}=\left({u}−\frac{\mathrm{1}}{{u}}\right)\:\:\:{dv}=\left(\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right){du} \\ $$$${so}\:\mathrm{2}{I}=\:{e}^{−\mathrm{2}} \int_{−\infty} ^{\infty} \:\:{e}^{−{v}^{\mathrm{2}} } {dv}={e}^{−\mathrm{2}} \sqrt{\pi} \\ $$$${finally}\:\:\:{I}=\:\frac{\sqrt{\pi}}{\mathrm{2}{e}^{\mathrm{2}} } \\ $$

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