$$\int_0^\infty e^{-(x-\frac{1}{x})^2} dx$$
Any hints?
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According to Wolfram Alpha, that integral is equal to $\frac{\sqrt{\pi}}{2}$. – mxian Apr 11 '17 at 18:32
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http://math.stackexchange.com/questions/106965/how-to-evaluate-the-integral-of-exp-x2-1-x2-on-0-infty – Hans Lundmark Apr 11 '17 at 18:38
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http://mathworld.wolfram.com/GlassersMasterTheorem.html – Jack D'Aurizio Apr 11 '17 at 20:35
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Hint. One may apply the following result, $$ \int_{-\infty}^{+\infty}f\left(x-\frac{a}x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x ,\qquad a>0, $$ which is true for any integrable $f$ over $\mathbb{R}$.
Olivier Oloa
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4The above hint gives a clear way:$$\int_0^\infty e^{-(x-\frac{1}{x})^2} dx=\frac12 \int_{-\infty}^\infty e^{-(x-\frac{1}{x})^2}dx =\frac12\int_{-\infty}^\infty e^{-x^2}dx. $$ – Olivier Oloa Apr 11 '17 at 18:35
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Putting $u=x-1/x$, $x = \frac{1}{2}u+\sqrt{4+u^2} $ so $dx = \left( 1 + \frac{u}{\sqrt{u^2+4}} \right)\frac{du}{2} $. The interval of integration changes to $(-\infty,\infty)$, so you have $$ \int_{-\infty}^{\infty} e^{-u^2}\left( 1 + \frac{u}{\sqrt{u^2+4}} \right)\frac{du}{2} = \frac{1}{2} \int_{-\infty}^{\infty} e^{-u^2} \, du $$ since the second term is odd and hence has integral zero. Of course, we know $\int_{-\infty}^{\infty} e^{-u^2} \, du=\sqrt{\pi}$.
Chappers
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$x = \frac{1}{y}$ : $$\int_0^\infty e^{-(x-\frac{1}{x})^2}dx = \int_0^\infty e^{-(\frac{1}{y}-y)^2}\frac{dy}{y^2}$$ $t =x-\frac{1}{x}, dt =(1+\frac{1}{x^2})dx $ : $$\int_0^\infty e^{-(x-\frac{1}{x})^2}dx = \frac12\int_0^\infty e^{-(x-\frac{1}{x})^2}(1+\frac{1}{x^2})dx = \frac12\int_{-\infty}^\infty e^{-t^2}dt = \frac{\sqrt{\pi}}{2}$$
reuns
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As Olivier Oloa mentioned, it works the same way with $\int_0^\infty f(x-\frac{1}{x}) dx = \frac{1}{2} \int_{-\infty}^\infty f(t)dt$ – reuns Apr 11 '17 at 20:30