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I have to solve $$ \int_{0} ^{\infty} e^{-x^{2} - \frac{t^{2}}{x^{2}}} dx $$

This is a standard integral in the treatment of improper integrals in many books but I haven't figured how to get the answer. The book I'm using (Elements of Real Analysis by Bartle) indicates to show that if $$ G(t)=\int_{0} ^{\infty} e^{-x^{2} - \frac{t^{2}}{x^{2}}} dx $$ then $G'(t)= 2G(t).$ Which I did very easily, but I don't know how that helps. I tried to use the standard techniques that involve computing the square of the integral with a change to polar coordinates but I couldn't make much progress with that. Any suggestions? Feel free to post a solution.

UPDATE Sorry, I made a little algebra mistake, this should be $G'(t)= -2G(t)$ instead, but the technique that's shown below still works, thanks for your answers.

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    $\begingroup$ If $G'(t) = 2 G(t)$, then $G(t) = A e^{2t}$.. $\endgroup$ – Cameron Williams May 5 at 16:53
  • $\begingroup$ @nicomezi Fair haha, but I'm not sure what else to say lol. $\endgroup$ – Cameron Williams May 5 at 16:54
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If $G'(t)=2G(t)$, then this is just a separable ODE. Hence, we get: \begin{equation} \begin{split} G'(t)=2G(t)\\ \frac{G'(t)}{G(t)}=2\\ \ln(G(t))=2t+C\\ G(t)=e^Ce^{2t}\\ G(0)=e^C=\int_0^\infty e^{-x^2}dx=\frac{\sqrt{\pi}}{2} \end{split} \end{equation}

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    $\begingroup$ You're missing a $\tfrac12$ factor. $\endgroup$ – J.G. May 5 at 17:08
  • $\begingroup$ @J.G. Thanks for this! $\endgroup$ – Andrew McMillan May 5 at 17:10
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Here's another technique that doesn't require the ODE. Since the integrand is even, the integral is $\tfrac12\int_{-\infty}^\infty e^{-(x-|t|/x)^2-2|t|}dx$. With Glasser's master theorem this becomes $\tfrac12\int_{-\infty}^\infty e^{-u^2-2|t|}du=\tfrac12\sqrt{\pi}e^{-2|t|}$.

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  • $\begingroup$ I also had Glasser's master theorem in mind. It's such an incredibly useful tool. $\endgroup$ – Cameron Williams May 5 at 17:11
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Solution is $G(t)=Ae^{-2t}$ Use $G(0)=\frac{\sqrt{\pi}}{2}$ to get $A$.

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  • $\begingroup$ In the original statement $G(t)=G(-t)$. $\endgroup$ – herb steinberg May 5 at 20:16

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