2.7 Improper integrals

In statistics, it is very common to encounter integrals that look like this:

$$\int_0^\infty f(x) \,dx.$$

This expression is a little confusing because if the integration region is infinite, then our earlier definition of the integral no longer works. What the expression means is that we’re taking the limit of the definite integrals (all of which are well-defined) as the region gets larger and larger:

$$\int_0^\infty f(x) \,dx= \lim_{b \to \infty} \int_a^b f(x) \,dx.$$

For example,

$$\int_0^b e^{-x} \,dx= 1 - e^{-b};$$

(if you don’t follow this derivation, see here and here). Since $\lim_{b \to \infty} e^{-b} = 0$ (see here), we have

$$\int_0^\infty e^{-x} \,dx= 1.$$

Integrals with an infinite bound (either upper or lower) are known as improper integrals. There are two other kinds of improper integrals.

Both bounds are infinite: One might expect that $\int_{-\infty}^\infty f(x) \,dx$ would be defined as the limit of $\int_{-a}^a f(x) \,dx$ as $a \to \infty$. But you’d be wrong! The actual definition is:

$$\int_{-\infty}^\infty f(x) \,dx= \int_{-\infty}^0 f(x) \,dx+ \int_0^\infty f(x) \,dx,$$

provided that both improper integrals exist. To see why, suppose we wanted to calculate $\int_{-\infty}^\infty x \,dx$. This integral does not exist, even though $\lim_{a \to \infty} \int_{-a}^a x \,dx= 0$. The problem with saying that $\int_{-\infty}^\infty x \,dx$ equals 0 is that it depends entirely on how fast the upper and lower bounds are going to infinity. For example, $\lim_{a \to \infty} \int_{-a}^{2a} x \,dx= \infty$. Without more information on exactly how fast the upper and lower bounds are going to infinity, $\int_{-\infty}^\infty x \,dx$ could equal anything.

Unbounded functions: For technical reasons, we also run into problems when $f$ is unbounded. Suppose we’re interested in integrating $f(x) = 1/\sqrt{x}$, for example. At $x=0$, $f(x)$ is undefined. Even if we were to define it, $f$ wouldn’t be continuous or monotone no matter what we chose, which causes problems with integration. As you might guess, however, we can extend our definition of the integral to include 0 as a lower bound by taking the limit as the bound goes to zero (if the limit exists):

$$\begin{align*} \int_0^1 x^{-1/2} \,dx&= \lim_{a \to 0} \int_a^1 x^{-1/2} \,dx\\ &= \lim_{a \to 0} 2\sqrt{x} \Big|_a^1 \\ &= 2\sqrt{1} - \lim_{a \to 0} 2\sqrt{a} \\ &= 2 \end{align*}$$

Note that in this case, if we failed to realize that $f$ was not bounded over the integration region and blindly plugged in 0 anyway, it wouldn’t make a difference – we’d get the same answer. However, this is not always true and if you ever run into a situation where this arises, it’s important to know the proper definition.