Gaussian Integral

$${\int }_{-\infty }^{\infty }{e}^{-x^2}dx=\sqrt{\pi }$$

Proof

$${\int }_{-\infty }^{\infty }{e}^{-x^2}dx{\int }_{-\infty }^{\infty }{e}^{-y^2}dx={\int }_{-\infty }^{\infty }{e}^{-x^2-y^2}dx$$

We convert to polar coordinates. So, $(x,y)=(r\cos \theta ,r\sin \theta )$ Note that the jacobian

$$|\frac{\partial (x,y)}{\partial (r,\theta )}|=r$$ $$⟹{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-x^2-y^2}dxdy={\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-1[(r\sin \theta {)}^2+(r\cos \theta {)}^2]}dxdy$$ $$={\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-r^2}dxdy=\pi$$