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# calculate-0-e-x-2-1-x-2-dx-

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