# How to compute the integral $\int_{-\infty}^\infty e^{-x^2}\,dx$? [duplicate]

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

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• – alexjo Nov 4 '13 at 23:54
• – Antonio Vargas Nov 5 '13 at 0:00

## 2 Answers

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.

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.

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