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.
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 \[I(\gamma) = \int_{-\infty}^\infty dx \; e^{-\gamma x^2} = \sqrt{\frac{\pi}{\gamma}}.\]