0

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

Harry Peter
  • 7,819
albmiz-mth
  • 303
  • 1
    See http://math.stackexchange.com/questions/429632/gaussian-integral – alexjo Nov 04 '13 at 23:54
  • 1
    See also http://math.stackexchange.com/q/105220/5531 and http://math.stackexchange.com/q/34767/5531 and http://math.stackexchange.com/q/390850/5531 . – Antonio Vargas Nov 05 '13 at 00:00

2 Answers2

3

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.

Cameron Buie
  • 102,994
2

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.

Pedro
  • 122,002
  • What allows one to bring the integrals together? Is it fubini's theorem or something else? – R R Nov 27 '13 at 03:58