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Math 220-2: Kursobjekt

Aaron Greicius

Section 1.24 Improper integrals

Definition 1.24.1. Improper integral of type I: infinite intervals.

Below we define definite integrals over infinite intervals. These are called improper integrals of type I, or integrals over infinite intervals.
  • Interval \([a,\infty)\).
    Let \(f\) be continuous on the interval \([a,\infty)\text{.}\) The integral of \(f\) over \([a,\infty)\), denoted \(\int_a^\infty f(x)\, dx\text{,}\) is said to converge if the limit
    \begin{equation} \lim_{R\to\infty}\int_a^R f(x)\, dx\tag{1.24.1} \end{equation}
    exists, in which case we define
    \begin{equation} \int_a^\infty f(x)\, dx=\lim_{R\to\infty}\int_a^R f(x)\, dx\text{.}\tag{1.24.2} \end{equation}
    If the limit (1.24.1) does not exist, we say the improper integral diverges.
  • Interval \((-\infty,a]\).
    Let \(f\) be continuous on the interval \((-\infty,a]\text{.}\) The integral of \(f\) over \((-\infty,a]\), denoted \(\int_{-\infty}^a f(x)\, dx\text{,}\) is said to converge if the limit
    \begin{equation} \lim_{R\to\infty}\int_{-R}^a f(x)\, dx\tag{1.24.3} \end{equation}
    exists, in which case we define
    \begin{equation} \int_{-\infty}^a f(x)\, dx=\lim_{R\to\infty}\int_{-R}^a f(x)\, dx\text{.}\tag{1.24.4} \end{equation}
    If the limit (1.24.3) does not exist, we say the improper integral diverges.
  • Real line.
    Let \(f\) be continuous on the real line \(\R=(-\infty,\infty)\text{.}\) The integral of \(f\) over \((-\infty,\infty)\), denoted \(\int_{-\infty}^{\infty}f(x)\, dx\text{,}\) is said to converge if both of the half-infinite integrals \(\int_{-\infty}^a f(x)\, dx\) and \(\int_a^{\infty}f(x)\, dx\) converge for some real number \(a\text{.}\) In this case we define
    \begin{equation} \int_{-\infty}^\infty f(x)\, dx=\int_{-\infty}^a f(x)\, dx+\int_a^{\infty}f(x)\, dx\text{.}\tag{1.24.5} \end{equation}
    If either (or both) of the half-infinite integrals diverge, we say that the integral of \(f\) over \((-\infty, \infty)\) diverges.

Example 1.24.3. Type I: half-infinite.

Evaluate \(\displaystyle\int_{-2}^{\infty}e^{-x}\, dx\text{.}\)
Solution.
We follow Procedure 1.24.2.
  1. Compute
    \begin{align*} I_R \amp = \int_{-2}^R e^{-x}\, dx\\ \amp = -e^{-x}\Bigr\vert_{-2}^R\\ \amp = e^2-e^{-R}\text{.} \end{align*}
  2. Compute
    \begin{align*} \lim_{R\to\infty} I_R \amp = \lim_{R\to\infty}e^2-e^{-R}\\ \amp = e^2- 0\\ \amp = e^2\text{.} \end{align*}
  3. We conclude that the improper integral converges and has value
    \begin{equation*} \int_{-2}^\infty e^{-x}\, dx= e^2\text{.} \end{equation*}

Example 1.24.4. Type I: \(p\)-test.

Evaluate \(\displaystyle\int_{2}^\infty \frac{1}{x^p}\, dx\) for \(p> 0\text{.}\)
Solution.
First consider
\begin{align*} I_R \amp = \int_2^R \frac{1}{x^p}\, dx\\ \amp = \begin{cases} \ln \abs{x} \Bigr\vert_2^R \amp \text{if } p=1 \\ \frac{1}{-p+1}x^{-p+1}\Bigr\vert_2^R \amp \text{if }p\ne 1 \end{cases}\\ \amp = \begin{cases} \ln R-\ln 2 \Bigr\vert_2^R \amp \text{if } p=1 \\ \frac{1}{-p+1}R^{-p+1}-\frac{1}{-p+1}2^{-p+1} \amp \text{if }p\ne 1 \end{cases}\text{.} \end{align*}
It follows that
\begin{align*} \lim_{R\to\infty}I_R \amp = \begin{cases} \lim_{R\to \infty} (\ln R-\ln 2) \amp \text{if } p=1 \\ \lim_{R\to \infty}\left( \frac{1}{-p+1}R^{-p+1}-\frac{1}{-p+1}2^{-p+1} \right) \amp \text{if }p\ne 1 \end{cases}\\ \amp = \begin{cases} \infty \amp \text{if } p=1 \\ \infty \amp \text{if } p< 1 \\ \frac{1}{(p-1)2^{p-1}} \amp p> 1 \end{cases}\text{.} \end{align*}
We conclude that
\begin{equation*} \int_2^\infty \frac{1}{x^p}\, dx =\begin{cases} \infty \amp \text{if } p\leq 1 \\ \frac{1}{(p-1)2^{p-1}} \amp \text{if } p> 0 \end{cases}\text{.} \end{equation*}
Simply by changing the lower limit of integration in Example 1.24.4 from \(2\) to \(a\text{,}\) we have essentially derived the type I statement of the \(p\)-test theorem.

Example 1.24.5. Type I: real line integral.

Evaluate \(\displaystyle\int_{-\infty}^{\infty}\frac{1}{x^2+1}\, dx\text{.}\)
Solution.
Let \(f(x)=\frac{1}{x^2+1}\text{.}\) Using Definition 1.24.1 we evaluate \(\int_{-\infty}^0 f(x)\, dx \) and \(\int_0^{\infty} f(x)\, dx\) separately. Observe that \(F(x)=\arctan x\) is an antiderivative of \(f(x)\text{.}\)
We have
\begin{align*} \int_{-\infty}^0f(x)\, dx \amp = \lim_{R\to \infty}\int_{-R}^0 f(x)\, dx\\ \amp = \lim_{R\to\infty} \arctan x\Bigr\vert_{-R}^0\\ \amp = \lim_{R\to \infty}\arctan(0)-\arctan(-R)\\ \amp = 0-\lim_{R\to\infty}\arctan(-R) \amp (\arctan 0=0)\\ \amp = -\left(-\frac{\pi}{2}\right) \amp (\lim_{x\to -\infty}\arctan x=-\frac{\pi}{2})\\ \amp =\frac{\pi}{2} \text{.} \end{align*}
Similarly, we have
\begin{align*} \int_{0}^{\infty}f(x)\, dx \amp = \lim_{R\to \infty}\int_{0}^R f(x)\, dx\\ \amp = \lim_{R\to\infty} \arctan x\Bigr\vert_{0}^R\\ \amp = \lim_{R\to \infty}\arctan(R)-\arctan(0)\\ \amp = 0-\lim_{R\to\infty}\arctan(R) \amp (\arctan 0=0)\\ \amp = \frac{\pi}{2} \amp (\lim_{x\to \infty}\arctan x=\frac{\pi}{2})\text{.} \end{align*}
Since both “half-infinite” improper integrals converge, we conclude that the given improper integral converges and has value
\begin{equation*} \int_{-\infty}^{\infty}f(x)\, dx=\int_{-\infty}^0f(x)\, dx+\int_0^\infty f(x)\, dx =\frac{\pi}{2}+\frac{\pi}{2}=\pi\text{.} \end{equation*}

Definition 1.24.6. Improper integrals of type II: discontinuities.

Assume \(f\) is continuous on the interval \(I=[a,b]\text{,}\) except possibly at one point.
  • Assume \(f\) is not continuous at \(x=a\text{.}\) The integral of \(f\) over \([a,b]\), denoted \(\int_a^b f(x)\, dx\text{,}\) is said to converge if the limit
    \begin{equation} \lim_{t\to a^+}\int_t^b f(x)\, dx\text{,}\tag{1.24.6} \end{equation}
    exists, in which case we define
    \begin{equation} \int_a^b f(x)\, dx=\lim_{t\to a^+}\int_t^b f(x)\, dx\text{.}\tag{1.24.7} \end{equation}
    If the limit (1.24.6) does not exist, we say the improper integral diverges.
  • Assume \(f\) is not continuous at \(x=b\text{.}\) The integral of \(f\) over \([a,b]\), denoted \(\int_a^b f(x)\, dx\text{,}\) is said to converge if the limit
    \begin{equation} \lim_{t\to b^-}\int_a^t f(x)\, dx\text{,}\tag{1.24.8} \end{equation}
    exists, in which case we define
    \begin{equation} \int_a^b f(x)\, dx=\lim_{t\to b^-}\int_a^t f(x)\, dx\text{.}\tag{1.24.9} \end{equation}
    If the limit (1.24.6) does not exist, we say the improper integral diverges.
  • Assume \(f\) is not continuous at \(c\in (a,b)\text{.}\) The integral of \(f\) over \([a,b]\), denoted \(\int_a^bf(x)\, dx\text{,}\) is said to converge if both improper integrals \(\int_a^cf(x)\, dx\) and \(\int_c^b f(x)\, dx\) converge, in which case we define
    \begin{equation} \int_a^bf(x)\, dx=\int_a^c f(x)\, dx+ \int_c^b f(x)\, dx,\text{.}\tag{1.24.10} \end{equation}
    Otherwise, we say that the integral over the entire interval diverges.

Definition 1.24.7. Area interpretation of improper integrals.

Let \(f\) be defined on an interval \(I\) for which the corresponding integral is improper, and let \(\mathcal{R}\) be the (potentially unbounded) region between the graph of \(f\) and the \(x\)-axis over the interval \(I\text{.}\)
  • We define the area (or total area) of \(\mathcal{R}\) to be the integral of \(\lvert f\rvert\) over \(I\text{,}\) assuming this integral converges.
  • We define the signed area of \(\mathcal{R}\) to be the integral of \(f\) over \(I\text{,}\) assuming this interval converges.

Example 1.24.9. Improper: type II.

Evaluate \(\displaystyle\int_{0}^2\frac{1}{x-1}\, dx\text{.}\)
Solution.
Let \(f(x)=\frac{1}{x-1}\text{.}\) Observe that the integral is improper because \(f\) is not defined at \(x=1\text{;}\) this is clearly the sole impropriety. Using definition Definition 1.24.6 we look at the improper integrals \(\int_0^1 f(x)\, dx\) and \(\int_1^2 f(x)\, dx\) separately. Observe that \(F(x)=\ln\abs{x-1}\) is an antiderivative of \(f(x)\text{.}\)
We have
\begin{align*} \int_0^1 f(x)\, dx \amp = \lim_{t\to 1^{-}}\int_0^t f(x)\, dx\\ \amp =\lim_{t\to 1^{-}}\ln\abs{t-1}-\ln \abs{-1}\\ \amp = -\infty-\ln 1 \amp (t\to 1^{-}\implies \abs{t-1}\to 0) \\ \amp = -\infty \end{align*}
Since \(\int_0^1 f(x)\, dx\) diverges, we conclude that \(\int_0^2 f(x)\, dx\) diverges, by definition.

Example 1.24.10. Improper: type II.

Evaluate \(\displaystyle\int_0^{1}\ln x\, dx\text{.}\)
Solution.
The integral is improper as \(\ln x\) is not defined at \(x=0\text{.}\) We follow Procedure 1.24.8.
  1. Compute
    \begin{align*} I_t \amp = \int_t^1\ln x\, dx\\ \amp = (x\ln x-x)\Bigr\vert_t^1\\ \amp = \ln 1-1-(t\ln t-t)\\ \amp = t-1-t\ln t\text{.} \end{align*}
  2. Compute
    \begin{align*} \lim_{t\to 0^+}I_t \amp = \lim_{t\to 0^+}t-1-t\ln t\\ \amp = -1-\lim_{t\to 0^+}\frac{\ln t}{1/t}\\ \amp = -1- \lim_{t\to 0^+}\frac{1/t}{-1/t^2}\\ \amp = -1+\lim_{t\to 0^+}t\\ \amp = -1\text{.} \end{align*}
  3. We conclude that the improper integral converges and has value
    \begin{equation*} \int_0^1\ln x\, dx=-1\text{.} \end{equation*}
    Geometrically, this tells us that the signed area of the region between the graph of \(y=\ln x\) and the \(x\)-axis, and between the lines \(x=0\) and \(x=1\) is \(-1\text{.}\) Since this region lies entirely below the \(x\)-axis, we see that its area is equal to 1.
Graph of natural log

Example 1.24.11. Improper: type II.

Evaluate \(\displaystyle\int_1^{4}\frac{x}{\sqrt[3]{x^2-4}}\, dx\text{.}\)
Solution.
Let \(f(x)=\frac{x}{\sqrt[3]{x^2-4}}\text{.}\) The integral is improper as the integrand is not defined at \(x=2\text{.}\) This is the only impropriety. Following Definition 1.24.6, we treat the improper integrals \(\int_1^2 f(x)\, dx\) and \(\int_2^4f(x)\, dx\) separately. Observe that \(F(x)=\frac{3}{4}(x^2-4)^{2/3}\) is an antiderivative of \(f(x)\text{.}\)
First compute
\begin{align*} \int_1^2f(x)\, dx \amp = \lim_{t\to 2^-}\int_1^t f(x)\, dx \\ \amp = \lim_{t\to 2^-}F(x)\Bigr\vert_1^t\\ \amp = \lim_{t\to 2^-}\frac{3}{4}((t^2-4)^{2/3}-(-3)^{2/3})\\ \amp = -\frac{3^{5/3}}{4}=3^{5/3}4^{-1}\text{.} \end{align*}
Next, we compute
\begin{align*} \int_2^4f(x)\, dx \amp = \lim_{t\to 2^+}\int_t^4 f(x)\, dx \\ \amp = \lim_{t\to 2^+}F(x)\Bigr\vert_t^4\\ \amp = \lim_{t\to 2^+}\frac{3}{4}(12^{2/3}-(t^2-4)^{2/3})\\ \amp = \frac{3^{5/3}}{4^{1/3}}=3^{5/3}4^{-1/3}\text{.} \end{align*}
Since both integrals converge, we conclude that \(\int_1^4f(x)\, dx\) converges and has value
\begin{equation*} \int_1^4f(x)\, dx=\int_1^2f(x)\, dx+\int_2^4 f(x)\, dx=-3^{5/3}4^{-1}+3^{5/3}4^{-1/3}\text{.} \end{equation*}
Geometrically, this result can be interpreted as the signed area of the region between the graph of \(f(x)\) and the \(x\)-axis, and between the lines \(x=1\) and \(x=4\text{.}\) Our analysis tells us that the areas of the two regions on either side of the vertical asymptote \(x=2\) are both finite. The diagram below then indicates that the signed area should be positive. Our integral computation is consistent with this since
\begin{equation*} -3^{5/3}4^{-1}+3^{5/3}4^{-1/3}=3^{5/3}4^{-1}(-1+4^{2/3})> 0\text{.} \end{equation*}
Graph of example integrand
Improper integrals (of both types) of power functions of the form \(1/x^p\) arise frequently enough to warrant their own theorem: the \(p\)-test theorem. As mentioned above the type I statement of the Theorem 1.24.12 was effectively proved in one of our earlier examples. The derivation of the type II statement is very similar.
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