Differential and Integral Calculus

Subsection 13.1 Definitions

The situation when integrating out to infinity is similar to the situation with infinite sums. Because there is no already assigned situation with infinite sums. Because there is no already assigned meaning for summing infinitely many things, we defined this as a limit, which in each case needs to be evaluated:

It is the same when one tries to integrate over the whole real line. We define such integrals by integrating over a bigger and bigger piece and taking the limit. In fact the definition is even pickier than that. We only let one of the limits of integration go to zero at a time. Consider first an integral over a half-line \([a,\infty)\text{.}\)

If we want both limits to be infinite then we require the two parts to be defined separately.

At this point you should be bothered by three questions.

What is \(c\text{?}\) Does it matter? How do you pick it?

The answer to the first question is, pick \(c\) to be anything, you'll always get the same answer. This is important because otherwise, what we wrote isn't really a definition. The reason the integral does not depend on \(c\) is that if one changes \(c\) from, say, 3 to 4, then the first of the two integrals loses a piece: \(\int_3^4 f(x) \, dx\text{.}\) But the second integral gains this same piece, so the sum is unchanged. This is true even if one or both pieces is infinite. Adding or subtracting the finite quantity \(\int_3^4 f(x) \, dx\) won't change that.

If we get \(-\infty + \infty\text{,}\) shouldn't that possibly be something other than "undefined"?

The answer to the second question is yes, sometimes you can be more specific. The one-sided integral to infinity is a limit. Cases where a finite limit does not exist can be resolved into limits of \(\infty\) or \(-\infty\text{,}\) along with the remaining cases where no limit exists even allowing for infinite limits. Because integrals over the whole real line are sums of one-sided (possibly infinite) limits, the rules for infinity from Sections~\ref{ss:variations} and~\ref{ss:LH} can be applied. In other words, integrals over the whole real line are the sum of two one-sided limits; we can add real numbers and \(\pm \infty\) according to the rules in Definition 7.1: \(\infty + \infty = \infty\) (and analogously with \(-\infty\)), \(\infty + a = \infty\) when \(a\) is real (and analogously with \(-\infty\)), \(\infty - \infty = UND\text{,}\) \(UND + {\rm anything } = UND\text{,}\) and so on.

Why do we have to split it up in the first place?

The third question is also a matter of definition. The reason we make the choice to do it this way is illustrated by the integral of the sign function

\begin{equation*} f(x) = {\rm sign} (x) = \left \{ \begin{array}{lr} 1 & x \geq 0 \\ 0 & x=0 \\ -1 & x \lt 0 \end{array} \right. \end{equation*}

On one hand, \(\int_{-M}^M f(x) \, dx\) is always zero, because the postive and negative parts exactly cancel. On the other hand, \(\int_M^\infty f(x) \, dx\) and \(\int_{-\infty}^M f(x) \, dx\) are always undefined. Do we want the answer for the whole integral \(\int_{-\infty}^\infty f(x) \, dx\) to be undefined or zero? There is no intrinsically correct choice here but it is a lot safer to have it undefined. If it has a value, one could make a case for values other than zero by centering the integral somewhere else, as in the following exercise.

Example 13.4.

The function \(\sin(x) / x\) is not defined at \(x=0\) but you might recall it does have a limit at 0, namely \(\lim_{x \to 0} \sin(x) / x = 1\text{.}\) Therefore the function

\begin{equation*} \operatorname{sinc} (x) := \begin{cases} \sin(x) / x & x \neq 0 \\ 1 & x=0 \end{cases} \end{equation*}

is a continuous function on the whole real line. Its graph is shown in Figure 13.5.

To write down a limit that defines this integral, we first choose any \(c\text{.}\) Choosing \(c=0\) makes things symmetric. The integral is then defined as the sum of two integrals, \(\int_{-\infty}^0 \operatorname{sinc}(x) \, dx + \int_0^{\infty} \operatorname{sinc}(x) \, dx\text{.}\) Going back to the definition of one-sided integrals as limits, this sum of integrals is equal to

\begin{equation*} \lim_{M \to -\infty} \int_M^0 \operatorname{sinc}(x) \, dx + \lim_{M \to \infty} \int_0^M \operatorname{sinc}(x) \, dx \, . \end{equation*}

It is not obvious whether these limits exist. One thing is easy to discern: because \(\operatorname{sinc}\) is an even function, the two limits have the same value (whether finite or not). We can safely say:

\begin{equation*} \int_{-\infty}^\infty \operatorname{sinc}(x) \, dx = 2 \cdot \lim_{M \to \infty} \int_0^M \operatorname{sinc} (x) \, dx \, . \end{equation*}

Subsection 13.2 Convergence

The central question of this section is: how do we tell whether a limit such as \(\int_b^\infty f(x) \, dx\) exists? If the limit exists, we would like to evaluate it if possible, or estimate it otherwise. When discussing convergence you should realize that \(\int_a^\infty f(x) \, dx\) either diverges for all values of \(a\) or converges for all values of \(a\) as long as \(f\) is defined and continuous on \([a,\infty)\text{.}\) For this reason, we use the notation \(\int^\infty f(x) \, dx\) or, to be really blunt, \(\displaystyle \int_{\rm who cares}^\infty f(x) \, dx\text{.}\)