Often it is necessary to integrate over the entire real line, or an unbounded interval of the form \([a,\infty)\) or \((-\infty,b]\text{.}\) We may also wish to integrate unbounded functions defined on an open bounded interval \((a,b)\text{.}\) For such intervals or functions, the Riemann integral is not defined, but we will write down the integral anyway in the spirit of Lemma 5.2.8. These integrals are called improper integrals and are limits of integrals rather than integrals themselves.
For a finite endpoint \(b\text{,}\) if \(f\) is bounded, then Lemma 5.2.8 says that we defined nothing new. What is new is that we can apply this definition to unbounded functions. The following set of examples is so useful that we state it as a proposition.
The proof follows by application of the fundamental theorem of calculus. Let us do the proof for \(p > 1\) for the infinite right endpoint and leave the rest to the reader. Hint: You should handle \(p=1\) separately.
We state the following proposition on “tails” for just one type of improper integral, though the proof is straightforward and the same for other types of improper integrals.
Let \(f \colon [a,\infty) \to \R\) be a function that is Riemann integrable on \([a,b]\) for all \(b > a\text{.}\) For every \(b > a\text{,}\) the integral \(\int_b^\infty f\) converges if and only if \(\int_a^\infty f\) converges, in which case
\begin{equation*}
\int_a^\infty f
=
\int_a^b f +
\int_b^\infty f .
\end{equation*}
Nonnegative functions are easier to work with as the following proposition demonstrates. The exercises will show that this proposition holds only for nonnegative functions. Analogues of this proposition exist for all the other types of improper limits and are left to the student.
Suppose \(f \colon [a,\infty) \to \R\) is nonnegative (\(f(x)
\geq 0\) for all \(x\)) and \(f\) is Riemann integrable on \([a,b]\) for all \(b > a\text{.}\)
Suppose \(\{ x_n \}_{n=1}^\infty\) is a sequence with \(\lim_{n\to\infty} x_n = \infty\text{.}\) Then \(\int_a^\infty f\) converges if and only if \(\lim_{n\to\infty} \int_a^{x_n} f\) exists, in which case
\begin{equation*}
\int_a^\infty f = \lim_{n\to\infty} \int_a^{x_n} f .
\end{equation*}
We start with the first item. As \(f\) is nonnegative, \(\int_a^x f\) is increasing as a function of \(x\text{.}\) If the supremum is infinite, then for every \(M \in \R\) we find \(N\) such that \(\int_a^N f \geq M\text{.}\) As \(\int_a^x f\) is increasing, \(\int_a^x f \geq M\) for all \(x \geq N\text{.}\) So \(\int_a^\infty f\) diverges to infinity.
Next suppose the supremum is finite, say \(A \coloneqq \sup \left\{ \int_a^x f : x \geq a \right\}\text{.}\) For every \(\epsilon > 0\text{,}\) we find an \(N\) such that \(A - \int_a^N f < \epsilon\text{.}\) As \(\int_a^x f\) is increasing, then \(A - \int_a^x f < \epsilon\) for all \(x \geq N\) and hence \(\int_a^\infty f\) converges to \(A\text{.}\)
Let us look at the second item. If \(\int_a^\infty f\) converges, then every sequence \(\{ x_n \}_{n=1}^\infty\) going to infinity works. The trick is proving the other direction. Suppose \(\{ x_n \}_{n=1}^\infty\) is such that \(\lim_{n\to\infty} x_n = \infty\) and
\begin{equation*}
\lim_{n\to\infty} \int_a^{x_n} f = A
\end{equation*}
converges. Given \(\epsilon > 0\text{,}\) pick \(N\) such that for all \(n \geq N\text{,}\) we have \(A - \epsilon < \int_a^{x_n} f < A + \epsilon\text{.}\) Because \(\int_a^x f\) is increasing as a function of \(x\text{,}\) we have that for all \(x \geq x_N\)
\begin{equation*}
A - \epsilon < \int_a^{x_N} f \leq \int_a^x f .
\end{equation*}
As \(\{ x_n \}_{n=1}^\infty\) goes to \(\infty\text{,}\) then for any given \(x\text{,}\) there is an \(x_m\) such that \(m \geq N\) and \(x \leq x_m\text{.}\) Then
\begin{equation*}
\int_a^{x} f \leq \int_a^{x_m} f < A + \epsilon .
\end{equation*}
In particular, for all \(x \geq x_N\text{,}\) we have \(\abs{\int_a^{x} f - A} < \epsilon\text{.}\)
Proposition5.5.5.Comparison test for improper integrals.
Let \(f \colon [a,\infty) \to \R\) and \(g \colon [a,\infty) \to \R\) be functions that are Riemann integrable on \([a,b]\) for all \(b > a\text{.}\) Suppose that for all \(x \geq a\text{,}\)
We start with the first item. For every \(b\) and \(c\text{,}\) such that \(a \leq b \leq c\text{,}\) we have \(-g(x) \leq f(x) \leq g(x)\text{,}\) and so
\begin{equation*}
\int_b^c -g \leq \int_b^c f \leq \int_b^c g .
\end{equation*}
In other words, \(\abs{\int_b^c f} \leq \int_b^c g\text{.}\)
\begin{equation*}
\int_a^\infty g =
\int_a^b g +
\int_b^\infty g .
\end{equation*}
As \(\int_a^b g\) goes to \(\int_a^\infty g\) as \(b\) goes to infinity, \(\int_b^\infty g\) goes to 0 as \(b\) goes to infinity. Choose \(B\) such that
\begin{equation*}
\int_B^\infty g < \epsilon .
\end{equation*}
As \(g\) is nonnegative, if \(B \leq b < c\text{,}\) then \(\int_b^c g < \epsilon\) as well. Let \(\{ x_n \}_{n=1}^\infty\) be a sequence going to infinity. Let \(M\) be such that \(x_n \geq B\) for all \(n \geq M\text{.}\) Take \(n, m \geq M\text{,}\) with \(x_n \leq x_m\text{,}\)
\begin{equation*}
\abs{\int_a^{x_m} f - \int_a^{x_n} f}
=
\abs{\int_{x_n}^{x_m} f}
\leq \int_{x_n}^{x_m} g < \epsilon .
\end{equation*}
Therefore, the sequence \(\bigl\{ \int_a^{x_n} f \bigr\}_{n=1}^\infty\) is Cauchy and hence converges.
We need to show that the limit is unique. Suppose \(\{ x_n \}_{n=1}^\infty\) is a sequence converging to infinity such that \(\bigl\{ \int_a^{x_n} f \bigr\}_{n=1}^\infty\) converges to \(L_1\text{,}\) and \(\{
y_n \}_{n=1}^\infty\) is a sequence converging to infinity such that \(\bigl\{ \int_a^{y_n} f \bigr\}_{n=1}^\infty\) converges to \(L_2\text{.}\) Then there must be some \(n\) such that \(\babs{\int_a^{x_n} f - L_1} < \epsilon\) and \(\babs{\int_a^{y_n} f - L_2} < \epsilon\text{.}\) We can also suppose \(x_n \geq B\) and \(y_n \geq B\text{.}\) Then
As \(\epsilon > 0\) was arbitrary, \(L_1 = L_2\text{,}\) and hence \(\int_a^\infty f\) converges. Above we have shown that \(\abs{\int_a^c f} \leq \int_a^c g\) for all \(c > a\text{.}\) By taking the limit \(c \to \infty\text{,}\) the first item is proved.
and try to integrate each part separately, you will not succeed. It is not true that you can split the improper integral in two; you cannot split the limit.
The last line in the computation does not even make sense. Both of the integrals diverge to infinity, since we can apply the comparison test appropriately with \(\nicefrac{1}{x}\text{.}\) We get \(\infty - \infty\text{.}\)
Suppose \(f \colon (a,b) \to \R\) is a function that is Riemann integrable on \([c,d]\) for all \(c\text{,}\)\(d\) such that \(a < c < d < b\text{,}\) then we define
\begin{equation*}
\int_a^b f \coloneqq \lim_{c \to a^+} \, \lim_{d \to b^-} \, \int_{c}^{d} f
\end{equation*}
One ought to always be careful about double limits. The definition given above says that we first take the limit as \(d\) goes to \(b\) or \(\infty\) for a fixed \(c\text{,}\) and then we take the limit in \(c\text{.}\) We will have to prove that in this case it does not matter which limit we compute first.
On the other hand, you must be careful to take the limits independently before you know convergence. Let \(f(x) = \frac{x}{\abs{x}}\) for \(x \not= 0\) and \(f(0) = 0\text{.}\) If \(a < 0\) and \(b > 0\text{,}\) then
\begin{equation*}
\int_{a}^b f
=
\int_{a}^0 f
+
\int_{0}^b f
=
a+b .
\end{equation*}
For every fixed \(a < 0\text{,}\) the limit as \(b \to \infty\) is infinite. So even the first limit does not exist, and the improper integral \(\int_{-\infty}^\infty f\) does not converge. On the other hand, if \(a > 0\text{,}\) then
\begin{equation*}
\int_{-a}^{a} f
=
(-a)+a = 0 .
\end{equation*}
Therefore,
\begin{equation*}
\lim_{a\to\infty}
\int_{-a}^{a} f
= 0 .
\end{equation*}
The integral of the sinc function is a continuous analogue of the alternating harmonic series \(\sum_{n=1}^\infty \nicefrac{{(-1)}^n}{n}\text{,}\) while the absolute value is like the regular harmonic series \(\sum_{n=1}^\infty \nicefrac{1}{n}\text{.}\) In particular, the fact that the integral converges must be done directly rather than using comparison test.
We will not prove the first statement exactly. Let us simply prove that the integral of the sinc function converges, but we will not worry about the exact limit. Because \(\frac{\sin(-x)}{-x} = \frac{\sin(x)}{x}\text{,}\) it is enough to show that
Figure5.9.Bound of \(\int_{2\pi n}^{2\pi (n+1)} \frac{\sin(x)}{x} \,dx\) using the shaded integral (signed area \(\frac{1}{\pi n}
+
\frac{-1}{\pi (n+1)}\)).
We find the partial sums of a series with positive terms. The series converges as \(\sum_{n=1}^\infty \frac{1}{\pi n (n+1)}\) is a convergent series. Thus as a sequence,
Let \(M > 2\pi\) be arbitrary, and let \(k \in \N\) be the largest integer such that \(2k\pi \leq M\text{.}\) For \(x \in [2k\pi,M]\text{,}\) we have \(\frac{-1}{2k\pi} \leq \frac{\sin(x)}{x} \leq \frac{1}{2k\pi}\text{,}\) and so
The double-sided integral of sinc also exists as noted above. We leave the other statement—that the integral of the absolute value of the sinc function diverges—as an exercise.
See Figure 5.10, for an illustration with \(k=1\text{.}\) By Proposition 5.2.11, \(f\) is integrable on every interval \([k,b]\) for all \(b > k\text{,}\) so the statement of the theorem makes sense without additional hypotheses of integrability.
Figure5.10.The area under the curve, \(\int_1^\infty f\text{,}\) is bounded below by the area of the shaded rectangles, \(f(2)+f(3)+f(4)+\cdots\text{,}\) and bounded above by the area entire rectangles, \(f(1)+f(2)+f(3)+\cdots\text{.}\)
Let \(\ell, m \in \Z\) be such that \(m > \ell \geq k\text{.}\) Because \(f\) is decreasing, we have \(\int_{n}^{n+1} f \leq f(n) \leq \int_{n-1}^{n} f\text{.}\) Therefore,
\begin{equation}
\int_\ell^m f
=
\sum_{n=\ell}^{m-1} \int_{n}^{n+1} f
\leq
\sum_{n=\ell}^{m-1} f(n)
\leq
f(\ell) +
\sum_{n=\ell+1}^{m-1} \int_{n-1}^{n} f
\leq
f(\ell)+
\int_\ell^{m-1} f .\tag{5.3}
\end{equation}
Suppose first that \(\int_k^\infty f\) converges and let \(\epsilon > 0\) be given. As before, since \(f\) is positive, then there exists an \(L \in \N\) such that if \(\ell \geq L\text{,}\) then \(\int_\ell^{m} f < \nicefrac{\epsilon}{2}\) for all \(m \geq \ell\text{.}\) The function \(f\) must decrease to zero (why?), so make \(L\) large enough so that for \(\ell \geq L\text{,}\) we have \(f(\ell) < \nicefrac{\epsilon}{2}\text{.}\) Thus, for \(m > \ell \geq L\text{,}\) we have via (5.3),
The series is therefore Cauchy and thus converges. The estimate in the proposition is obtained by letting \(m\) go to infinity in (5.3) with \(\ell = k\text{.}\)
Conversely, suppose \(\int_k^\infty f\) diverges. As \(f\) is positive, then by Proposition 5.5.4, the sequence \(\{ \int_k^m f \}_{m=k}^\infty\) diverges to infinity. Using (5.3) with \(\ell = k\text{,}\) we find
\begin{equation*}
\int_k^m f
\leq
\sum_{n=k}^{m-1} f(n) .
\end{equation*}
As the left-hand side goes to infinity as \(m \to \infty\text{,}\) so does the right-hand side.
The integral test can be used not only to show that a series converges, but to estimate its sum to arbitrary precision. Let us show \(\sum_{n=1}^\infty \frac{1}{n^2}\) exists and estimate its sum to within 0.01. As this series is the \(p\)-series for \(p=2\text{,}\) we already proved it converges (let us pretend we do not know that), but we only roughly estimated its sum.
In other words, \(\nicefrac{1}{k} + \sum_{n=1}^{k-1} \nicefrac{1}{n^2}\) is an estimate for the sum to within \(\nicefrac{1}{k^2}\text{.}\) Therefore, if we wish to find the sum to within 0.01, we note \(\nicefrac{1}{{10}^2} = 0.01\text{.}\) We obtain
Take \(f \colon [0,\infty) \to \R\text{,}\) Riemann integrable on every interval \([0,b]\text{,}\) and such that there exist \(M\text{,}\)\(a\text{,}\) and \(T\text{,}\) such that \(\abs{f(t)} \leq M e^{at}\) for all \(t \geq T\text{.}\) Show that the Laplace transform of \(f\) exists. That is, for every \(s > a\) the following integral converges:
Let \(f \colon \R \to \R\) be a Riemann integrable function on every interval \([a,b]\text{,}\) and such that \(\int_{-\infty}^\infty \abs{f(x)}\,dx < \infty\text{.}\) Show that the Fourier sine and cosine transforms exist. That is, for every \(\omega \geq 0\) the following integrals converge
Suppose \(f \colon [0,\infty) \to \R\) is Riemann integrable on every interval \([0,b]\text{.}\) Show that \(\int_0^\infty f\) converges if and only if for every \(\epsilon > 0\) there exists an \(M\) such that if \(M \leq a < b\text{,}\) then \(\babs{\int_a^b f} < \epsilon\text{.}\)
Find an example of an unbounded continuous function \(f \colon
[0,\infty) \to \R\) that is nonnegative and such that \(\int_0^\infty f < \infty\text{.}\) Note that \(\lim_{x\to\infty} f(x)\) will not exist; compare previous exercise. Hint: On each interval \([k,k+1]\text{,}\)\(k \in \N\text{,}\) define a function whose integral over this interval is less than say \(2^{-k}\text{.}\)
(More challenging) Find an example of a function \(f \colon [0,\infty) \to \R\) integrable on all intervals such that \(\lim_{n\to\infty} \int_0^n f\) converges as a limit of a sequence (so \(n \in \N\)), but such that \(\int_0^\infty f\) does not exist. Hint: For all \(n\in \N\text{,}\) divide \([n,n+1]\) into two halves. On one half make the function negative, on the other make the function positive.
It is sometimes desirable to assign a value to integrals that normally cannot be interpreted even as improper integrals, e.g. \(\int_{-1}^1 \nicefrac{1}{x}\,dx\text{.}\) Suppose \(f \colon [a,b] \to \R\) is a function and \(a < c < b\text{,}\) where \(f\) is Riemann integrable on the intervals \([a,c-\epsilon]\) and \([c+\epsilon,b]\) for all \(\epsilon > 0\text{.}\) Define the Cauchy principal value of \(\int_a^b f\) as
\begin{equation*}
p.v.\!\int_a^b f \coloneqq \lim_{\epsilon\to 0^+}
\left(
\int_a^{c-\epsilon} f +
\int_{c+\epsilon}^b f
\right) ,
\end{equation*}
Compute \(\lim_{\epsilon\to 0^+}
( \int_{-1}^{-\epsilon} \nicefrac{1}{x}\,dx +
\int_{2\epsilon}^1 \nicefrac{1}{x}\,dx )\) and show it is not equal to the principal value.
Suppose \(f \colon [-1,1] \to \R\) is an odd function (\(f(-x)=-f(x)\)) that is integrable on \([-1,-\epsilon]\) and \([\epsilon,1]\) for all \(\epsilon >0\text{.}\) Prove that \(p.v.\!\int_{-1}^1 f = 0\)
Let \(f \colon \R \to \R\) and \(g \colon \R \to \R\) be continuous functions, where \(g(x) = 0\) for all \(x \notin [a,b]\) for some interval \([a,b]\text{.}\)
