KindiakMath

The Gaussian Integral

Problem 1. Evaluate the Gaussian integral \displaystyle \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx.

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Solution. Since e^{-x^2} is even in x,

\displaystyle \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx = 2 \cdot \int_{0}^{\infty} e^{-x^2}\, \mathrm dx.

For the right-hand side,

\begin{aligned}\left( \int_{0}^{\infty} e^{-x^2}\, \mathrm dx \right)^2 &= \left( \int_{0}^{\infty} e^{-x^2}\, \mathrm dx \right) \cdot \left( \int_{0}^{\infty} e^{-y^2}\, \mathrm dy \right) \\ &= \int_0^\infty e^{-x^2} \cdot \left( \int_0^\infty e^{-y^2}\, \mathrm dy \right)\, \mathrm dx \\ &= \int_0^\infty e^{-x^2} \cdot \left( \int_0^\infty e^{-u^2x^2} \cdot x\, \mathrm du \right)\, \mathrm dx \\ &= \int_0^\infty \int_0^\infty xe^{-x^2(u^2+1)} \, \mathrm du\, \mathrm dx. \end{aligned}

By Fubini’s theorem,

\begin{aligned}\left( \int_{0}^{\infty} e^{-x^2}\, \mathrm dx \right)^2 &= \int_0^\infty \int_0^\infty xe^{-x^2(u^2+1)} \, \mathrm du\, \mathrm dx\\ &= \int_0^\infty \int_0^\infty xe^{-x^2(u^2+1)} \, \mathrm dx\, \mathrm du \\ &= \int_0^\infty \left[ -\frac 1{2(u^2+1)} \cdot e^{-x^2(u^2+1)} \right]_0^\infty\, \mathrm du \\ &= \int_0^\infty \frac 1{2(u^2+1)}\, \mathrm du \\ &= \frac 12 \cdot [\tan^{-1}(u)]_0^{\infty} \\ &= \frac 12 \cdot \frac{\pi}2 = \frac{\pi}{4}. \end{aligned}

Taking square roots,

\displaystyle \int_{0}^{\infty} e^{-x^2}\, \mathrm dx = \frac{\sqrt{\pi}}{2} \quad \Rightarrow \quad \int_{-\infty}^{\infty} e^{-x^2}\, \mathrm dx = \sqrt{\pi}.

This calculation was authored by Hirokazu Iwasawa.

—Joel Kindiak, 26 Jul 25, 2156H

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