Question

The value of the integral \(I=\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x\) is (a) Use polar coordinates to evaluate the improper integral $$ \begin{aligned} I^{2} &=\left(\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x\right)\left(\int_{-\infty}^{\infty} e^{-y^{2} / 2} d y\right) \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right) / 2} d A \end{aligned} $$ (b) Use the result of part (a) to determine \(I\). For more information on this problem, see the article "Integrating \(e^{-x^{2}}\) Without Polar Coordinates" by William Dunham in Mathematics Teacher. To view this article, go to the website ww.matharticles.com

Short answer

The value of the integral \(I=\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x\) is \(\sqrt{2\pi}\)

Step-by-step solution

Convert to a double integral

Square both sides, which allows to treat the original integral as a double integral in the x-y plane: \(I^{2} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right) / 2} dx dy \)

Convert to polar coordinates

In polar coordinates, \(x^{2}+y^{2}=r^{2}\) and \(dx dy = r dr d\theta\). Also, the limits for \(r\) and \(\theta\) are from \(0\) to \(\infty\) and from \(0\) to \(2\pi\) respectively. So, \(I^{2} = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^{2}/2} r dr d\theta \)

Evaluate the improper integrals

Separate out the integrals and evaluate each. For \(\theta\) integral, \( \int_{0}^{2\pi} d\theta = 2\pi \). For \(r\) integral, use the substitution \(u = r^{2}/2\), \(du = r dr\) which leads to \( \int_{0}^{\infty} e^{-u} du = 1 \). So, \(I^2 = 2\pi \cdot 1 \) or \(I^2 = 2\pi\)

Solve for I

Since \(I\) is defined for \(r\geq0\), and \(\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x\) is positive, take only the positive root to get \(I = \sqrt{2\pi}\)