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How to compute the integral $\int_{-\infty}^\infty e^{-x^2}\,dx$ using polar coordinates?

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marked as duplicate by Antonio Vargas, Daniel Robert-Nicoud, Stefan4024, user7530, Hanul Jeon Nov 5 '13 at 0:43

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Hint: Let $I=\int_{-\infty}^\infty e^{-x^2}\,dx.$ Then $$I^2=\left(\int_{-\infty}^\infty e^{-x^2}\,dx\right)\left(\int_{-\infty}^\infty e^{-y^2}\,dy\right)=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^2}e^{-y^2}\,dx\,dy=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}\,dx\,dy.$$ Now switch to polar coordinates.

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The usual tear-free approach is to write $$\left(\int_{-\infty}^\infty e^{-x^2}\,dx\right)^2=\int_{-\infty}^\infty e^{-x^2}\,dx\int_{-\infty}^\infty e^{-y^2}\,dy=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-(x^2+y^2)}\,dx\,dy$$ and change to polar coordinates.

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  • $\begingroup$ What allows one to bring the integrals together? Is it fubini's theorem or something else? $\endgroup$ – R R Nov 27 '13 at 3:58

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