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Question Number 136197 by Ar Brandon last updated on 19/Mar/21

$$\mathrm{a}.\mathrm{Prove}\mathrm{that}\mathrm{for}\mathrm{any}\mathrm{real}\mathrm{constant}\mathrm{a}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{\mathrm{a}}{{\mathrm{x}}^2}}\mathrm{dx}=\mathrm{\infty }$$$$\mathrm{b}.\mathrm{If}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}\mathrm{real}\mathrm{constants},\mathrm{explain}\mathrm{why}\mathrm{we}\mathrm{cannot}\mathrm{split}\mathrm{the}$$$$\mathrm{integral}{\int }_0^{\mathrm{\infty }}({\mathrm{e}}^{-\frac{\mathrm{a}}{{\mathrm{x}}^2}}-{\mathrm{e}}^{-\frac{\mathrm{b}}{{\mathrm{x}}^2}})\mathrm{dx}\mathrm{as}\mathrm{the}\mathrm{difference}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{\mathrm{a}}{{\mathrm{x}}^2}}\mathrm{dx}-{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{\mathrm{b}}{{\mathrm{x}}^2}}\mathrm{dx}$$$$\mathrm{c}.\mathrm{If}\mathrm{a}⩾0\mathrm{and}\mathrm{b}⩾0\mathrm{constants},\mathrm{then}\mathrm{prove}\mathrm{that}$$$${\int }_0^{\mathrm{\infty }}({\mathrm{e}}^{-\frac{\mathrm{a}}{{\mathrm{x}}^2}}-{\mathrm{e}}^{-\frac{\mathrm{b}}{{\mathrm{x}}^2}})\mathrm{dx}=\sqrt{\pi \mathrm{b}}-\sqrt{\pi \mathrm{a}}.$$$$\mathrm{d}.\mathrm{If}\mathrm{a}>\mathrm{b}⩾0\mathrm{constants},\mathrm{then}\mathrm{prove}\mathrm{that}{\int }_0^{\mathrm{\infty }}({\mathrm{e}}^{-\frac{\mathrm{a}}{{\mathrm{x}}^2}}-{\mathrm{e}}^{-\frac{\mathrm{b}}{{\mathrm{x}}^2}})\mathrm{dx}=\mathrm{\infty }$$

Answered by Dwaipayan Shikari last updated on 19/Mar/21

$$I\left(a\right){\int }_0^{\mathrm{\infty }}(e^{-\frac{a}{x^2}}-e^{-\frac{b}{x^2}})dx$$$$I^{\prime }\left(a\right)=-{\int }_0^{\mathrm{\infty }}\frac{e^{-\frac{a}{x^2}}}{x^2}dx\frac{1}{x}=u\Rightarrow -\frac{1}{x^2}=\frac{du}{dx}$$$$=-{\int }_0^{\mathrm{\infty }}{e}^{-{au}^2}du=-\frac{1}{2}\sqrt{\frac{\pi }{a}}$$$$I\left(a\right)=-\sqrt{\pi a}+C\Rightarrow I\left(b\right)=-\sqrt{\pi b}+C=0\Rightarrow C=\sqrt{\pi b}$$$$I\left(a\right)=\sqrt{\pi }(\sqrt{b}-\sqrt{a})$$

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