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
$\endgroup$ 43 Answers
$\begingroup$\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}
$\endgroup$ 1 $\begingroup$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 $y = g(x)$.
$\endgroup$ $\begingroup$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 $0$?
$\endgroup$
