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Given $$\int_0^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$$ evaluate: $$\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx. $$

I can find that $$\left(ax+\frac{b}{x}\right)^2 = a^2x^2+2ab+\frac{b^2}{x^2}$$

therefore: $$\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx = e^{2ab}\int_0^{\infty}e^{-\left(ax+\frac{b}{x}\right)^2}dx$$

but I can't find any clue then.

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  • $\begingroup$ I think that setting $t = ax + b/x $ and using integration by parts two times should work. Note also that the minus should be outside the square in the exponent $\endgroup$
    – Ant
    Jun 9, 2016 at 13:25

1 Answer 1

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Assuming $a,b>0$, you have: $$ I= e^{-2ab}\int_{0}^{+\infty}e^{-(ax-b/x)^2}\,dx = e^{-2ab}\int_{-\infty}^{+\infty}\frac{1+\color{blue}{\frac{z}{\sqrt{4ab+z^2}}}}{2a}\,e^{-z^2}\,dz $$ through the substitution $ax-\frac{b}{x}=z$, but the contribute given by the blue term vanishes by symmetry, hence:

$$ I = \color{red}{\frac{\sqrt{\pi}}{2a\, e^{2ab}}}.$$

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  • $\begingroup$ Neat trick! It does implicitly assume $a> 0$ and $b > 0$ for the limits of integration to work, though. I think the general result for general $a$ and $b$ can be obtained by taking $a \to |a|$ and $b \to |b|$. $\endgroup$ Jun 9, 2016 at 13:38
  • $\begingroup$ @MichaelSeifert: absolutely right. $\endgroup$ Jun 9, 2016 at 13:39

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