Gaussian Integrals

This page is about Gaussian Integrals, that is integrals of the form

\begin{align} \int_{-\infty}^{\infty} e^{-x^2}dx \end{align}

Needless to say, there are many variations of this integral and often certain integrals can be related to an integral of this form. The applications of such integrals are widespread in science.

We begin with the evaluation of the first integral encountered here, denoting it by I. First of all, a change of the independent variable obviously doesn't affect the value of I (it is obvious from the Riemann criterion for integrability that the above integral converges). Hence we have

\begin{align} I = \int_{-\infty}^{\infty} e^{-x^2}dx = \int_{-\infty}^{\infty} e^{-y^2}dy \end{align}

This allows us to write

\begin{align} I^{2} = \int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy \end{align}

which can be rewritten as

\begin{align} I^{2} = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}dxdy \end{align}

We now switch to polar coordinates $(r,\theta)$ from $(x,y)$ by using the following substitutions:

\begin{align} x = r\cos\theta \qquad y = r\sin\theta \end{align}

This necessitates the introduction of the Jacobian, that is,

\begin{align} dxdy = \left|\begin{array}{cc}{\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta}\\\frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta}\end{array}\right|drd\theta = rdrd\theta \end{align}

Therefore we have

\begin{align} I^{2} = \int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}rdrd\theta = \pi \end{align}

and hence

\begin{align} \int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi} \end{align}

As a first observation, the integrand in (1) is an even function and hence, the integral over the positive half line is half the integral over the real line. That is

\begin{align} \int_{0}^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2} \end{align}

In contrast, there is no way we can compute the value of this integral if the limits are both finite. Only approximate values or series solutions are available in that case.

What about integrands of the form $f(x)e^{-x^2}$ where f(x) is a continuous function over the whole real line? Consider an integral of this form, given by

\begin{align} \int_{-\infty}^{\infty}f(x)e^{-x^2}dx \end{align}

It is obvious that such an integral will converge provided f(x) does not go to infinity faster than $e^{-x^2}$ goes to zero as $x \rightarrow \pm \infty$. Suppose f(x) is a polynomial of degree n (n > 0). By linearity of the integral, it is sufficient to consider the integral

\begin{align} \int_{-\infty}^{\infty} x^{n}e^{-x^2}dx \end{align}

We observe that if n is odd, then the integrand is an odd function and hence

\begin{align} \int_{-\infty}^{\infty} x^{n}e^{-x^2}dx = \lim_{a \rightarrow \infty}\int_{-a}^{a} x^{n}e^{-x^2}dx = 0 \end{align}

For even n = 2m, the integral is

\begin{align} \int_{-\infty}^{\infty} x^{2m}e^{-x^2}dx = 2\int_{0}^{\infty} x^{2m}e^{-x^2}dx \end{align}

Now, one way to compute this integral which is particularly appealing is to consider the integral,

\begin{align} I(a) = \int_{-\infty}^{\infty}e^{-x^2/a^2}dx = a\int_{-\infty}^{\infty}e^{-x^2}dx = a\sqrt{\pi} \end{align}

as a function of the continuous real variable a. Indeed,

\begin{align} I'(a) = \int_{-\infty}^{\infty}\frac{2x^2}{a^3}e^{-x^2/a^2}dx = a\int_{-\infty}^{\infty}e^{-x^2}dx = a\sqrt{\pi} \end{align}

and hence,

\begin{align} \int_{-\infty}^{\infty}x^{2}e^{-x^2/a^2}dx = \frac{a^4}{2}\sqrt{\pi} \end{align}

Differentiating both sides wrt a, we get

\begin{align} I''(a) = \int_{-\infty}^{\infty}\frac{2x^4}{a^3}e^{-x^2/a^2}dx = 2a^{3}\sqrt{\pi} \end{align}

and hence

\begin{align} \int_{-\infty}^{\infty}x^{4}e^{-x^2/a^2}dx = a^{6}\sqrt{\pi} \end{align}

and so on. We can do better if we try to get some kind of a recurrence relation (edit) and the final answer works out to be

\begin{align} \int_{-\infty}^{\infty}x^{2m}e^{-x^2/a^2}dx = \frac{(2m-1)!!}{2^m}a^{2m+1}\sqrt{\pi} \end{align}

where $(2m-1)!!$ denotes the double factorial, defined by

\begin{align} (2m-1)!! = \Pi_{k = 0}^{k=2m-2}}(2m-1-k) \end{align}
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