3.5: The Gaussian Integral

Here’s a famous integral:

$$\int_{-\infty}^\infty \; e^{-\gamma x^2} \; dx.$$ The integrand is called a Gaussian, or bell curve, and is plotted below. The larger the value of $\gamma$, the more narrowly-peaked the curve.

Figure $\PageIndex{1}$

The integral was solved by Gauss in a brilliant way. Let $I(\gamma)$ denote the value of the integral. Then $I^2$ is just two independent copies of the integral, multiplied together:

$$I^2(\gamma) = \left[\int_{-\infty}^\infty dx\; e^{-\gamma x^2}\right] \times \left[\int_{-\infty}^\infty dy\; e^{-\gamma y^2}\right].$$ Note that in the second copy of the integral, we have changed the “dummy” label $x$ (the integration variable) into $y$, to avoid ambiguity. Now, this becomes a two-dimensional integral, taken over the entire 2D plane: $$I^2(\gamma) = \int_{-\infty}^\infty dx\, \int_{-\infty}^\infty dy \; e^{-\gamma (x^2+y^2)}.$$ Next, change from Cartesian to polar coordinates: $$I^2(\gamma) = \int_{0}^\infty dr\, r \int_{0}^{2\pi} d\phi \; e^{-\gamma r^2} = \left[ \int_{0}^\infty dr\, r \, e^{-\gamma r^2}\right] \times \left[\int_{0}^{2\pi} d\phi \right] = \frac{1}{2\gamma} \cdot 2\pi.$$ By taking the square root, we arrive at the result $$I(\gamma) = \int_{-\infty}^\infty dx \; e^{-\gamma x^2} = \sqrt{\frac{\pi}{\gamma}}.$$