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Question Number 118188 by mnjuly1970 last updated on 16/Oct/20

Commented by MJS_new last updated on 16/Oct/20

$$f\left(x\right)={\mathrm{e}}^{-\frac{4}{x^2}}$$$$\underset{x\to 0}{\lim }f\left(x\right)=0$$$$\underset{x\to \pm \mathrm{\infty }}{\lim }f\left(x\right)=1$$$$f^{\prime }\left(x\right)>0\mathrm{\forall }x>0$$$$\Rightarrow \mathrm{integral}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{converge}$$

Commented by mnjuly1970 last updated on 16/Oct/20

$$youareright.$$$$icorrectedit.$$$$thankyou..$$

Answered by aleks041103 last updated on 16/Oct/20

$$u=\frac{1}{x}\Rightarrow x=\frac{1}{u}\Rightarrow dx=-\frac{1}{u^2}du$$$$\Rightarrow \int e^{-1/x^2}dx=-\int e^{-u^2}\frac{du}{u^2}=\int e^{-u^2}d\left(\frac{1}{u}\right)$$$$IBP$$$$\int e^{-u^2}d\left(\frac{1}{u}\right)=\frac{e^{-u^2}}{u}-\int \frac{1}{u}d\left(e^{-u^2}\right)=$$$$=\frac{e^{-u^2}}{u}+\int 2{ue}^{-u^2}\frac{1}{u}du=$$$$=\sqrt{\pi }erf\left(u\right)+\frac{e^{-u^2}}{u}=\sqrt{\pi }erf\left(1/x\right)+{xe}^{-1/x^2}$$$$\Rightarrow \int e^{-1/x^2}dx=\sqrt{\pi }erf\left(1/x\right)+{xe}^{-1/x^2}$$$$\int e^{-{\left(a/x\right)}^2}dx=a\int e^{-{\left(1/\left(x/a\right)\right)}^2}d\left(x/a\right)=$$$$=[a\sqrt{\pi }erf\left(\frac{a}{x}\right)+{xe}^{-{\left(a/x\right)}^2}]$$$${\int }_0^{\mathrm{\infty }}{e}^{-{\left(a/x\right)}^2}dx=-{[a\sqrt{\pi }erf\left(\frac{a}{x}\right)+{xe}^{-{\left(a/x\right)}^2}]}_{\mathrm{\infty }}^0=$$$$=-a\sqrt{\pi }+{\underset{x\to \mathrm{\infty }}{limxe}}^{-{\left(a/x\right)}^2}$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}(e^{-\frac{4}{x^2}}-e^{-\frac{9}{x^2}})dx=$$$$={\int }_0^{\mathrm{\infty }}{e}^{-{\left(2/x\right)}^2}dx-{\int }_0^{\mathrm{\infty }}{e}^{-{\left(3/x\right)}^2}dx=$$$$=-2\sqrt{\pi }+3\sqrt{\pi }-\underset{x\to \mathrm{\infty }}{lim}\left[x(e^{-{\left(3/x\right)}^2}-e^{-{\left(2/x\right)}^2})\right]=$$$$=\sqrt{\pi }-\underset{x\to \mathrm{\infty }}{lim}\left[x((1-\frac{9}{x^2})-(1-\frac{4}{x^2}))\right]$$$$=\sqrt{\pi }+5\underset{x\to \mathrm{\infty }}{lim}\left(\frac{1}{x}\right)=\sqrt{\pi }$$$$\mathrm{\Omega }=\sqrt{\pi }$$

Commented by mnjuly1970 last updated on 16/Oct/20

$$bravo$$$$thankyousomuch..$$

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