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How do I transform the integral $$\int_0^\infty e^{-x^2} dx$$ from 0 to $\infty$ to o to 1 and. I have to devise a monte carlo algorithm to solve this further, so any advise would be of great help

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Are you aware that this integral can be solved exactly? –  Alex Becker Apr 13 '13 at 19:38
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If you would like to evaluate this integral, that may not be the best way to go. This is called the Gaussian integral, and its value is $\sqrt{\pi}/2$. en.wikipedia.org/wiki/Gaussian_integral –  Lord Soth Apr 13 '13 at 19:39
    
The idea is that $$\int_0^\infty e^{-x^2}dx=\frac12\sqrt{\int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy}=\frac12\sqrt{\int_0^{2\pi}\int_0^\infty re^{-r^2}drd\theta}=\frac12\sqrt{\pi}$$ –  Alex Becker Apr 13 '13 at 19:46
    
Thanks a lot for this, true can be solved straight away, but have to devise a MC based solution for the same, any thoughts would be of great help –  sleepybob Apr 13 '13 at 20:01

3 Answers 3

\begin{align} \int_0^{\infty} e^{-x^2}dx & = \int_0^{1} e^{-x^2}dx + \underbrace{\int_1^{\infty} e^{-x^2}dx}_{x \mapsto 1/x} = \int_0^1 e^{-x^2}dx + \int_1^0 e^{-1/x^2} \left(\dfrac{-dx}{x^2} \right)\\ &=\int_0^1 \left(e^{-x^2} + \dfrac{e^{-1/x^2}}{x^2}\right)dx \end{align}

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could you provide some background on the transformation from x to 1/x as the limits goes from 1 to 0, after the transformation. Please could you advice –  sleepybob Apr 13 '13 at 20:00

Pick your favorite invertible, increasing function $f : (0,1) \to (0,+\infty)$. Make a change of variable $x = f(y)$.

Or, pick your favorite invertible, increasing function $g : (0,+\infty) \to (0,1)$. Make a change of variable $y = g(x)$.

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This is a pretty weird integral. There is no elementary antiderivative for $e^{-x^2}$. However, the area under the entire curve is $\sqrt{\pi}$. The details escape me at the moment, but you can derive this be rotating the curve around the y axis, changing to polar, then integrating. So, if $$\int_{-\infty}^\infty e^{-x^2} = \sqrt{\pi}$$ what happens when the lower bound is $0$?

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