$Evaluate the integral $\int_{-\infty}^{+\infty}e^{-x^2}dx$$

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$Evaluate the integral $\int_{-\infty}^{+\infty}e^{-x^2}dx$$

Is there a way to use double integrals to evaluate the following integral?
$$\int_{-\infty}^{+\infty}e^{-x^2}dx$$

Calculus Multivariable Calculus Integrals

Melissa G

46

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Let $A=\int_{-\infty}^{+\infty}e^{-x^2}dx$, and note also that $A=\int_{-\infty}^{+\infty}e^{-y^2}dy$. Hence
\[A^2=\int_{-\infty}^{+\infty}e^{-x^2}dx \times\int_{-\infty}^{+\infty}e^{-y^2}dy=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy.\]
Thus using polar coordinates we have
$$A^2=\int_{0}^{\infty}\int_0^{2\pi}e^{-r^2}rd\theta dr=2\pi\int_{0}^{\infty}e^{-r^2}r dr=\frac{2\pi}{-2}\int_{0}^{\infty}e^{-r^2}(-2r)$$
\[=-\pi e^{-r^2}\mid_0^{\infty}=-\pi(0-1)=\pi.\]
Hence
\[A^2=\pi \Rightarrow A=\sqrt{\pi},\]
and therefore
\[\int_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt{\pi}.\]

Daniel

53

$Evaluate the integral $\int_{-\infty}^{+\infty}e^{-x^2}dx$$

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