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Question Number 98423 by mathmax by abdo last updated on 13/Jun/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^2-\frac{1}{{\mathrm{x}}^2}}\mathrm{dx}$$

Answered by maths mind last updated on 14/Jun/20

$$={\int }_0^{\mathrm{\infty }}{e}^{-{(x-\frac{1}{x})}^2-2}dx...t=\frac{1}{x}\Rightarrow$$$$={\int }_0^{+\mathrm{\infty }}{e}^{-{(t-\frac{1}{t})}^2-2}\frac{dt}{t^2}$$$$\Rightarrow 2{\int }_0^{+\mathrm{\infty }}{e}^{-x^2-\frac{1}{x^2}}dx=e^{-2}{\int }_0^{+\mathrm{\infty }}(1+\frac{1}{t^2})e^{-{(t-\frac{1}{t})}^2}$$$$u=t-\frac{1}{t}$$$$=e^{-2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{e}^{-u^2}du=e^{-2}.\sqrt{\pi }$$$$\Rightarrow {\int }_0^{+\mathrm{\infty }}{e}^{-x^2-\frac{1}{x^2}}dx=\frac{\sqrt{\pi }}{2e^2}$$$$$$

Commented by abdomathmax last updated on 14/Jun/20

$$\mathrm{thanks}\mathrm{sir}.$$

Commented by maths mind last updated on 14/Jun/20

$$withepleasur$$

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