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Question Number 136476 by Ar Brandon last updated on 22/Mar/21

$$\left(\mathrm{a}\right)\mathrm{Let}$$$$\mathrm{I}\left(\alpha \right)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{(\mathrm{x}-\frac{\alpha }{\mathrm{x}})}^2}\mathrm{dx}$$$$\mathrm{Show}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{legitimate}\mathrm{to}\mathrm{take}\mathrm{the}\mathrm{derivative}\mathrm{of}\mathrm{I}\left(\alpha \right)\mathrm{and}\mathrm{also}{\mathrm{I}}^{\prime }\left(\alpha \right)=$$$$0.\mathrm{Then}\mathrm{show}\mathrm{that}$$$$\mathrm{I}\left(\alpha \right)=\frac{\sqrt{\pi }}{2}.$$$$\left(\mathrm{b}\right)\mathrm{Use}\left(\mathrm{a}\right)\mathrm{to}\mathrm{prove}$$$${\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-({\mathrm{x}}^2+{\alpha }^2{\mathrm{x}}^{-2})}\mathrm{dx}=\frac{\sqrt{\pi }}{2}{\mathrm{e}}^{-2\alpha }.$$$$$$

Answered by Dwaipayan Shikari last updated on 22/Mar/21

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

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