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A long while ago (at university) I learned the following (fairly standard) method for evaluating $I:=\int_{-\infty}^{\infty}e^{-z^2}dz$ by squaring $I$ then using polar coordinates:

$$ \begin{align} I^2 &= \int_{-\infty}^{\infty}e^{-z^2}dz \int_{-\infty}^{\infty}e^{-z^2}dz \\ &= 4\int_{0}^{\infty}e^{-x^2}dx \int_{0}^{\infty}e^{-y^2}dy \\ &= 4\int_{0}^{\infty}\int_{0}^{\infty}e^{-x^2}e^{-y^2}dxdy \\ &= 4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(x^2+y^2)}dxdy \\ &= 4\int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}e^{-r^2}rdrd\theta \\ &= 2\pi\int_{0}^{\infty}re^{-r^2}dr \\ &= \pi \left[ -e^{-r^2} \right]_{0}^{\infty} \\ &= \pi \end{align} \\ \\ \implies I=\sqrt{\pi} $$

I remember being amazed at the ingenuity of that first line: squaring $I$ in order to turn the single integral into a double integral.

I am curious whether any other methods are known for evaluating $I$. Especially methods that do not begin by squaring $I$.

I have not been able to do anything worthwhile with it myself. My research has yielded this method that does not require polar coordinates but still begins by squaring $I$. I have also considered complex integration using a clever contour but have not managed to complete a method.

Does anybody know of any other methods?

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marked as duplicate by Jyrki Lahtonen, mathcounterexamples.net, Community Aug 9 '18 at 10:32

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Another way is to integrate $$\int_0^\infty\int_0^\infty x\exp(-x^2(1+y^2))\,dx\,dy= \int_0^\infty\int_0^\infty x\exp(-x^2(1+y^2))\,dy\,dx$$ both ways. The first gives $$\frac12\int_0^\infty\frac{dy}{1+y^2}=\frac\pi4$$ and the second gives $$\left(\int_0^\infty \exp(-t^2)\,dt \right)^2.$$

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